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Calculus 3 : Multi-Variable Chain Rule

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Example Questions

Example Question #1313 : Calculus 3

Compute $\frac{dz}{dt}$ $\displaystyle \frac{dz}{dt}$ for $z=2xe^{xy}$ $\displaystyle z=2xe^{xy}$, x=t^3 $\displaystyle x=t^3$, $y=t^{-2}$ $\displaystyle y=t^{-2}$.

Possible Answers:

$\frac{dz}{dt}=6t^2e^{xy}-4t^{-3}x^2e^{xy}$ $\displaystyle \frac{dz}{dt}=6t^2e^{xy}-4t^{-3}x^2e^{xy}$

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$ $\displaystyle \frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(6t^2e^{xy}-2xye^{xy})-(4t^{-3}x^2e^{xy})$ $\displaystyle \frac{dz}{dt}=(6t^2e^{xy}-2xye^{xy})-(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(4t^{-3}x^2e^{xy})$ $\displaystyle \frac{dz}{dt}=(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})$ $\displaystyle \frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})$

Correct answer:

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$ $\displaystyle \frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

Explanation:

All we need to do is use the formula for multivariable chain rule.

$\frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$ $\displaystyle \frac{dz}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$

$\frac{\partial f}{\partial x}=2e^{xy}+2xye^{xy}$ $\displaystyle \frac{\partial f}{\partial x}=2e^{xy}+2xye^{xy}$

$\frac{dx}{dt}=3t^2$ $\displaystyle \frac{dx}{dt}=3t^2$

$\frac{\partial f}{\partial y}=2x^2e^{xy}$ $\displaystyle \frac{\partial f}{\partial y}=2x^2e^{xy}$

$\frac{dy}{dt}=-2t^{-3}$ $\displaystyle \frac{dy}{dt}=-2t^{-3}$

When we put this all together, we get.

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$ $\displaystyle \frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

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Example Question #2 : Multi Variable Chain Rule

Evaluate $\frac{\partial a}{\partial x}$ $\displaystyle \frac{\partial a}{\partial x}$ in terms of $\displaystyle r$ and/or $\displaystyle s$ if $\displaystyle a= x-3y+z^2$, x=r^3 $\displaystyle x=r^3$, y= s+2r $\displaystyle y= s+2r$, and z=6r $\displaystyle z=6r$.

Possible Answers:

$\displaystyle 3r^2-3s-6r+12r^2$

$\displaystyle 3r^2+72r-6$

$\displaystyle 1$

$\displaystyle 3r^2+72rs-6$

$\displaystyle 3r^6 +72r-6$

Correct answer:

$\displaystyle 3r^2+72r-6$

Explanation:

Expand the equation for chain rule.

$\frac{\partial a}{\partial x} = \frac{\partial a}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial a}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial a}{\partial z}\frac{\partial z}{\partial r}$ $\displaystyle \frac{\partial a}{\partial x} = \frac{\partial a}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial a}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial a}{\partial z}\frac{\partial z}{\partial r}$

Evaluate each partial derivative.

$\frac{\partial a}{\partial x} = 1$ $\displaystyle \frac{\partial a}{\partial x} = 1$, $\frac{\partial x}{\partial r} =3r^2$ $\displaystyle \frac{\partial x}{\partial r} =3r^2$

$\frac{\partial a}{\partial y}= -3$ $\displaystyle \frac{\partial a}{\partial y}= -3$, $\frac{\partial y}{\partial r} = 2$ $\displaystyle \frac{\partial y}{\partial r} = 2$

$\frac{\partial a}{\partial z} =2z$ $\displaystyle \frac{\partial a}{\partial z} =2z$, $\frac{\partial z}{\partial r} = 6$ $\displaystyle \frac{\partial z}{\partial r} = 6$

Substitute the terms in the equation.

$\frac{\partial a}{\partial x} =(1)(3r^2)+(-3)(2)+(2z)(6) = 3r^2-6+12z$ $\displaystyle \frac{\partial a}{\partial x} =(1)(3r^2)+(-3)(2)+(2z)(6) = 3r^2-6+12z$

Substitute z=6r $\displaystyle z=6r$.

$\displaystyle 3r^2-6+12(6r) = 3r^2+72r-6$

The answer is: $\displaystyle 3r^2+72r-6$

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Example Question #2 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $\displaystyle r=z^2-2y+1$, z=t^2 $\displaystyle z=t^2$, $\displaystyle y=t^3-2t$.

Possible Answers:

$\frac{dr}{dt}=9-t$ $\displaystyle \frac{dr}{dt}=9-t$

$\frac{dr}{dt}=-t^3+5t^2+2t-3$ $\displaystyle \frac{dr}{dt}=-t^3+5t^2+2t-3$

$\frac{dr}{dt}=12t^2+3t-7$ $\displaystyle \frac{dr}{dt}=12t^2+3t-7$

$\frac{dr}{dt}=4t^3-6t^2+4$ $\displaystyle \frac{dr}{dt}=4t^3-6t^2+4$

Correct answer:

$\frac{dr}{dt}=4t^3-6t^2+4$ $\displaystyle \frac{dr}{dt}=4t^3-6t^2+4$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle z$ and $\displaystyle y$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial z}\frac{dz}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$ $\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial z}\frac{dz}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$=\left ( 2z\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$ $\displaystyle =\left ( 2z\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

$=\left ( 2t^2\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$ $\displaystyle =\left ( 2t^2\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

$\displaystyle =4t^3-6t^2+4$

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Example Question #4 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $\displaystyle r=3a^4+5b^2$, $\displaystyle a=t^2-6t+12$, b=t-5 $\displaystyle b=t-5$.

Possible Answers:

$\frac{dr}{dt}=t^7-4t^6-52t^5-25t^4+88t^3-186t^2+2281t-1266$ $\displaystyle \frac{dr}{dt}=t^7-4t^6-52t^5-25t^4+88t^3-186t^2+2281t-1266$

$\frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$ $\displaystyle \frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

$\frac{dr}{dt}=4t^3-6t^2+4$ $\displaystyle \frac{dr}{dt}=4t^3-6t^2+4$

$\frac{dr}{dt}=-54t^6+52t^5-250t^4+828t^3-1624t^2+2206t-1246$ $\displaystyle \frac{dr}{dt}=-54t^6+52t^5-250t^4+828t^3-1624t^2+2206t-1246$

Correct answer:

$\frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$ $\displaystyle \frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle a$ and $\displaystyle b$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$ $\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$=\left ( 12a^3\right )\left ( 2t-6\right )+\left ( 10b\right )\left ( 1\right )$ $\displaystyle =\left ( 12a^3\right )\left ( 2t-6\right )+\left ( 10b\right )\left ( 1\right )$

$=\left ( 12\left ( t^2-6t+12\right )^3\right )\left ( 2t-6\right )+\left ( 10\left ( t-5\right )\right )\left ( 1\right )$ $\displaystyle =\left ( 12\left ( t^2-6t+12\right )^3\right )\left ( 2t-6\right )+\left ( 10\left ( t-5\right )\right )\left ( 1\right )$

$=\left12 ( t^6-18t^5+144t^4-648t^3+1728t^2-2592t+1728\right )\left ( 2t-6\right )+10t-50$ $\displaystyle =\left12 ( t^6-18t^5+144t^4-648t^3+1728t^2-2592t+1728\right )\left ( 2t-6\right )+10t-50$

$\displaystyle =24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228096t-124416 +10t-50$

$\displaystyle =24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

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Example Question #5 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $\displaystyle r=cos(ab)$, $\displaystyle a=t^3+7t$, $\displaystyle b=5t^2-12t+1$.

Possible Answers:

$\frac{dr}{dt}=4t^3-6t^2+4sin\left ( 4t^3-6t^2+4 \right )$ $\displaystyle \frac{dr}{dt}=4t^3-6t^2+4sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(t^4-3t^3-128t^2-49t)sin(t^5-131t^4+66t^3-98t^2+17t-121)$ $\displaystyle \frac{dr}{dt}=(t^4-3t^3-128t^2-49t)sin(t^5-131t^4+66t^3-98t^2+17t-121)$

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$ $\displaystyle \frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

$\frac{dr}{dt}=(5t^4+8t^3-18t^2-9t)cos(4t^5-11t^4+6t^3-8t^2+7t-11)$ $\displaystyle \frac{dr}{dt}=(5t^4+8t^3-18t^2-9t)cos(4t^5-11t^4+6t^3-8t^2+7t-11)$

Correct answer:

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$ $\displaystyle \frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle a$ and $\displaystyle b$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\displaystyle =b*cos(ab)(3t^2+7)+a*cos(ab)(10t-12)$

$=-(5t^2-12t+1)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 3t^2+7\right )-(t^3+7t)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 10t-12\right )$ $\displaystyle =-(5t^2-12t+1)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 3t^2+7\right )-(t^3+7t)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 10t-12\right )$

$\displaystyle =-(15^4-36t^3+38t^2-84t+7)sin(5t^5-12t^4+36t^3-84t^2+7t)-(10t^4-12t^3+70t^2-84t)sin(5t^5-12t^4+36t^3-84t^2+7t)$

$\displaystyle =(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

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Example Question #6 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $\displaystyle r=sin(ab^2)$, a=5t-1 $\displaystyle a=5t-1$, b=-9t^3 $\displaystyle b=-9t^3$.

Possible Answers:

$\frac{dr}{dt}=(4t^3-6t^2+4)sin\left ( 4t^3-6t^2+4 \right )$ $\displaystyle \frac{dr}{dt}=(4t^3-6t^2+4)sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$ $\displaystyle \frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(-90t^4+18t^3)cos(405t^7-81t^6)$ $\displaystyle \frac{dr}{dt}=(-90t^4+18t^3)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$ $\displaystyle \frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$

Correct answer:

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$ $\displaystyle \frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle a$ and $\displaystyle b$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{\partial r}{\partial a}=cos(ab^2)(b^2)$ $\displaystyle \frac{\partial r}{\partial a}=cos(ab^2)(b^2)$

$\displaystyle =cos(405t^7-81t^6)(81t^6)$

$\frac{da}{dt}=5$ $\displaystyle \frac{da}{dt}=5$

$\frac{\partial r}{\partial b}=cos(ab^2)(2ab)$ $\displaystyle \frac{\partial r}{\partial b}=cos(ab^2)(2ab)$

$\displaystyle =cos(405t^7-81t^6)(-90t^4+18t^3)$

$\frac{db}{dt}=-27t^2$ $\displaystyle \frac{db}{dt}=-27t^2$

$\frac{dr}{dt}=\left ( cos(405t^7-81t^6)(81t^6) \right )(5)+(cos(405t^7-81t^6)(-90t^4+18t^3))(-27t^2)$ $\displaystyle \frac{dr}{dt}=\left ( cos(405t^7-81t^6)(81t^6) \right )(5)+(cos(405t^7-81t^6)(-90t^4+18t^3))(-27t^2)$

$\displaystyle =405t^6*cos(405t^7-81t^6)+(2430t^6-486t^5)cos(405t^7-81t^6)$

$\displaystyle =(2745t^6-486t^5)cos(405t^7-81t^6)$

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Example Question #7 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $r=e^{x+y}$ $\displaystyle r=e^{x+y}$, $\displaystyle x=5t+t^2$, y=7-7t $\displaystyle y=7-7t$.

Possible Answers:

$\frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$ $\displaystyle \frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$

$\frac{dr}{dt}=(2t-2)e^{t^2-2t+7}$ $\displaystyle \frac{dr}{dt}=(2t-2)e^{t^2-2t+7}$

$\frac{dr}{dt}=e^{t^2-2t+7}$ $\displaystyle \frac{dr}{dt}=e^{t^2-2t+7}$

$\frac{dr}{dt}=5te^{x+y}$ $\displaystyle \frac{dr}{dt}=5te^{x+y}$

Correct answer:

$\frac{dr}{dt}=(2t-2)e^{t^2-2t+7}$ $\displaystyle \frac{dr}{dt}=(2t-2)e^{t^2-2t+7}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle x$ and $\displaystyle y$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$ $\displaystyle \frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=e^{x+y}$ $\displaystyle \frac{\partial r}{\partial x}=e^{x+y}$

$\frac{dx}{dt}=5+2t$ $\displaystyle \frac{dx}{dt}=5+2t$

$\frac{\partial r}{\partial y}=e^{x+y}$ $\displaystyle \frac{\partial r}{\partial y}=e^{x+y}$

$\frac{dy}{dt}=-7$ $\displaystyle \frac{dy}{dt}=-7$

$\frac{dr}{dt}=e^{x+y}(5+2t)+e^{x+y}(-7)$ $\displaystyle \frac{dr}{dt}=e^{x+y}(5+2t)+e^{x+y}(-7)$

$\frac{dr}{dt}=(2t-2)e^{x+y}$ $\displaystyle \frac{dr}{dt}=(2t-2)e^{x+y}$

$\frac{dr}{dt}=(2t-2)e^{5t+t^2+7-7t}=(2t-2)e^{t^2-2t+7}$ $\displaystyle \frac{dr}{dt}=(2t-2)e^{5t+t^2+7-7t}=(2t-2)e^{t^2-2t+7}$

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Example Question #361 : Partial Derivatives

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $r=e^{xy}$ $\displaystyle r=e^{xy}$, $\displaystyle x=-2t^3+1$, $\displaystyle y=t^2-t-1$.

Possible Answers:

$\frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$ $\displaystyle \frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

$\frac{dr}{dt}=e^{t^2-2t+7}$ $\displaystyle \frac{dr}{dt}=e^{t^2-2t+7}$

$\frac{dr}{dt}=(-9t^4+23t^3+t^2-5t-1)e^{(-2t^3+2t^2-t-1)}$ $\displaystyle \frac{dr}{dt}=(-9t^4+23t^3+t^2-5t-1)e^{(-2t^3+2t^2-t-1)}$

$\frac{dr}{dt}=(t^2+7t-6)e^{(3t^2-9t-11)}$ $\displaystyle \frac{dr}{dt}=(t^2+7t-6)e^{(3t^2-9t-11)}$

Correct answer:

$\frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$ $\displaystyle \frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle x$ and $\displaystyle y$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{\partial r}{\partial x}=ye^{xy}=(t^2-t-1)e^{xy}$ $\displaystyle \frac{\partial r}{\partial x}=ye^{xy}=(t^2-t-1)e^{xy}$

$\frac{dx}{dt}=-6t^2$ $\displaystyle \frac{dx}{dt}=-6t^2$

$\frac{\partial r}{\partial y}=xe^{xy}=(-2t^3+1)e^{xy}$ $\displaystyle \frac{\partial r}{\partial y}=xe^{xy}=(-2t^3+1)e^{xy}$

$\frac{dy}{dt}=2t-1$ $\displaystyle \frac{dy}{dt}=2t-1$

$\frac{dr}{dt}=(t^2-t-1)e^{xy}(-6t^2)+(-2t^3+1)e^{xy}(2t-1)$ $\displaystyle \frac{dr}{dt}=(t^2-t-1)e^{xy}(-6t^2)+(-2t^3+1)e^{xy}(2t-1)$

$=(-6t^4+12t^3+6t^2)e^{xy}+(-4t^4+2t^3+2t-1)e^{xy}$ $\displaystyle =(-6t^4+12t^3+6t^2)e^{xy}+(-4t^4+2t^3+2t-1)e^{xy}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{xy}$ $\displaystyle =(-10t^4+14t^3+6t^2+2t-1)e^{xy}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^3+1)(t^2-t-1)}$ $\displaystyle =(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^3+1)(t^2-t-1)}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$ $\displaystyle =(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

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Example Question #4 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $\displaystyle r=ln(ab)$, $\displaystyle a=-2t^2-3t$, b=4t+9 $\displaystyle b=4t+9$.

Possible Answers:

$\frac{dr}{dt}=\frac{3t^3-11t^2+7t-14}{t^4-t^3-7t^2+7t+1}$ $\displaystyle \frac{dr}{dt}=\frac{3t^3-11t^2+7t-14}{t^4-t^3-7t^2+7t+1}$

$\frac{dr}{dt}=\frac{2t^2-6t+7}{-t^3+3t^2+7t}$ $\displaystyle \frac{dr}{dt}=\frac{2t^2-6t+7}{-t^3+3t^2+7t}$

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$ $\displaystyle \frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

$\frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$ $\displaystyle \frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

Correct answer:

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$ $\displaystyle \frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle a$ and $\displaystyle b$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{\partial r}{\partial a}=\frac{b}{ab}=\frac{1}{a} =\frac{1}{-2t^2-3t}$ $\displaystyle \frac{\partial r}{\partial a}=\frac{b}{ab}=\frac{1}{a} =\frac{1}{-2t^2-3t}$

$\frac{da}{dt}=-4t-3$ $\displaystyle \frac{da}{dt}=-4t-3$

$\frac{\partial r}{\partial b}=\frac{a}{ab}=\frac{1}{b}=\frac{1}{4t+9}$ $\displaystyle \frac{\partial r}{\partial b}=\frac{a}{ab}=\frac{1}{b}=\frac{1}{4t+9}$

$\frac{db}{dt}=4$ $\displaystyle \frac{db}{dt}=4$

$\frac{dr}{dt}=\left ( \frac{1}{-2t^2-3t} \right )(-4t-3)+\left ( \frac{1}{4t+9} \right )(4)$ $\displaystyle \frac{dr}{dt}=\left ( \frac{1}{-2t^2-3t} \right )(-4t-3)+\left ( \frac{1}{4t+9} \right )(4)$

$=\frac{-4t-3}{-2t^2-3t}+\frac{4}{4t+9}$ $\displaystyle =\frac{-4t-3}{-2t^2-3t}+\frac{4}{4t+9}$

$=\frac{(-4t-3)(4t+9)+4(-2t^2-3t)}{(-2t^2-3t)(4t+9)}$ $\displaystyle =\frac{(-4t-3)(4t+9)+4(-2t^2-3t)}{(-2t^2-3t)(4t+9)}$

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$ $\displaystyle \frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

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Example Question #5 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ when $\displaystyle r=ln(5a+3b)$, a=8t-3 $\displaystyle a=8t-3$, $\displaystyle b=t^2-7t+1$.

Possible Answers:

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$ $\displaystyle \frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

$\frac{dr}{dt}=\frac{-t^2-9t+11}{t^2-9t+4}$ $\displaystyle \frac{dr}{dt}=\frac{-t^2-9t+11}{t^2-9t+4}$

$\frac{dr}{dt}=\frac{t-4}{t^2-9t+12}$ $\displaystyle \frac{dr}{dt}=\frac{t-4}{t^2-9t+12}$

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$ $\displaystyle \frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

Correct answer:

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$ $\displaystyle \frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ $\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$.

Since $\displaystyle a$ and $\displaystyle b$ are both functions of $\displaystyle t$, $\frac{dr}{dt}$ $\displaystyle \frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{\partial r}{\partial a}=\frac{5}{5a+3b}=\frac{5}{5(8t-3)+3(t^2-7t+1)}=\frac{5}{3t^2+19t-12}$ $\displaystyle \frac{\partial r}{\partial a}=\frac{5}{5a+3b}=\frac{5}{5(8t-3)+3(t^2-7t+1)}=\frac{5}{3t^2+19t-12}$

$\frac{da}{dt}=8$ $\displaystyle \frac{da}{dt}=8$

$\frac{\partial r}{\partial b}=\frac{3}{5a+3b}=\frac{3}{5(8t-3)+3(t^2-7t+1)}=\frac{3}{3t^2+19t-12}$ $\displaystyle \frac{\partial r}{\partial b}=\frac{3}{5a+3b}=\frac{3}{5(8t-3)+3(t^2-7t+1)}=\frac{3}{3t^2+19t-12}$

$\frac{db}{dt}=2t-7$ $\displaystyle \frac{db}{dt}=2t-7$

$\frac{dr}{dt}=\left ( \frac{5}{3t^2+19t-12} \right )(8)+\left ( \frac{3}{3t^2+19t-12} \right )(2t-7)$ $\displaystyle \frac{dr}{dt}=\left ( \frac{5}{3t^2+19t-12} \right )(8)+\left ( \frac{3}{3t^2+19t-12} \right )(2t-7)$

$=\left ( \frac{40}{3t^2+19t-12} \right )+\left ( \frac{6t-21}{3t^2+19t-12} \right )$ $\displaystyle =\left ( \frac{40}{3t^2+19t-12} \right )+\left ( \frac{6t-21}{3t^2+19t-12} \right )$

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$ $\displaystyle \frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

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Calculus 3 : Multi-Variable Chain Rule

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