How to find differential functions - AP Calculus AB

Card 0 of 12012

Question

Find the derivative of $f(x)=\sqrt{1-\cos^{2}(x)}$

Answer

To solve this problem, we need the derivative of a constant, the derivative of the trigonometric function cosine, and the chain rule.

First, let's rewrite the function in terms of a power:

$f(x)=(1-\cos^{2}(x))^{\frac{1}{2}}$

Now we should apply the chain rule which states that:

$\frac{d}{dx} h(i(x))=h'(i(x))i'(x)$

In this problem, $h(x)=(1-\cos^{2}(x))^{\frac{1}{2}}$ and $i(x)=1-\cos^{2}(x)$ .

To find we need to use the power rule, which states:

$\frac{d}{dx}(x^{n})=nx^{n-1}$

$h'(x)=\frac{1}{2}(1-\cos^{2}(x))^{\frac{1}{2}-1}=\frac{1}{2}(1-\cos^{2}(x))^{-\frac{1}{2}}=\frac{1}{2\sqrt{1-\cos^{2}(x)}}$

To find , we again need to use the chain rule, the derivative of a constant, and the derivative of the rtigonometric function cosine to evaluate $\frac{d}{dx}\cos^{2}(x)$ , which state that:

$\frac{d}{dx} \cos(x)= -\sin(x), \frac{d}{dx}(c)=0$

$i'(x)=2\cos(x)sin(x)$

Plugging all of these equations back into the chain rule, we obtain:

$f'(x)=\frac{1}{2}(1-\cos^{2}(x))^{-\frac{1}{2}}*2\sin(x)\cos(x)$

$\frac{\sin(x)\cos(x)}{\sqrt{1-cos^{2}(x)}}$

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Question

Differentiate:

$f(x) = \left(\frac{1}{x}\right)\cos^2(x)$

Answer

To find the derivative of this function we must use the Product Rule and the Chain Rule. First we set

$h(x) = \frac{1}{x}$

and

$g(x) = \cos^2(x)$

Now differentiating both of these functions gives

$h'(x) = \frac{-1}{x^2}$

$g'(x) = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x)$

Applying this to the Product Rule gives us,

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

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Question

Differentiate the function

$f(x) = \sin(\cos(2x))$

Answer

To differentiate the function properly, we must use the Chain Rule which is,

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))g'(x)$

Therefore the derivative of the function is,

$f'(x) = \cos(\cos(2x))(-\sin(2x))(2) = -2\cos(\cos(2x))\sin(2x)$

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Question

Differentiate

$f(x) = \sin^2(x^2)$

Answer

To differentiate this equation we use the Chain Rule.

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Using this throughout the equation gives us,

$f'(x) = 2\sin(x^2)\cos(x^2)(2x)$

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Question

Find the derivative of

$f(x) = 4\sin(x+3x^2)$

Answer

To find the derivative of the function we must use the Chain Rule, which is

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we get,

$f'(x) = 4\cos(x+3x^2)(1+6x)$

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Question

Determine the slope of the line that is normal to the function $f(x)=cos^3(2\pi x^3)$ at the point $x=\frac{1}{2}$

Answer

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=cos^3(2\pi x^3)$ at the point $x=\frac{1}{2}$

The slope of the tangent is

$f'(x)=-18\pi x^2sin(2\pi x^3)cos^2(2\pi x^3)$

$f'(\frac{1}{2})=-9\pi (\frac{1}{2})^2sin(2\pi (\frac{1}{2})^3)cos^2(2\pi (\frac{1}{2})^3)=\frac{-9\sqrt{2}\pi}{16}$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$m=-\frac{1}{\frac{-9\sqrt{2}\pi}{16}}=\frac{8\sqrt{2}}{9\pi}$

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Question

Find the derivative of

$f(x) =8\sin(4x+3)$

Answer

To find the derivative of the function we must use the Chain Rule

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we are given gives,

$f'(x) = 8\cos(4x+3)(4) = 32\cos(4x+3)$

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Question

Differentiate the function:

Answer

To differentiate this problem we will need to use the power rule.

The power rule is, where n is the exponent.

Thus our derivative is,

$f'(z)=\frac{d}{dz}(az^-^7)=-7\cdot az^-^7^-^1=-7az^-^8$ .

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Question

Find the first derivative of .

Answer

We need to differentiate term by term, applying the power rule,

$(x^n)'=nx^{n-1}$

This gives us

$\begin{align} (2x^3-6x+2)' &= 6x^2-6 \end{align}$

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Question

Differentiate the function:

$f(x)=-12x^\frac{-3}{4}$

Answer

To find the derivative of this function we will need to use the power rule.

The power rule is where n is the exponent.

Therefore, our derivative is

$f'(x)=\left(-\frac{3}{4}\right)\cdot (-12)x^\frac{-3}{4}^-^{1}=9x^{\frac{-7}{4}}$ .

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Question

Find the derivative.

Answer

Use the power rule to find this derivative.

$\frac{d}{dx}5x^2=10x$

$\frac{d}{dx}3x=3$

Recall that the derivative of a constant is zero.

Thus, the derivative is .

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Question

Find the derivative.

$\frac{4}{x^3}$

Answer

Use the quotient rule to find this derivative.

$\frac{(x^3)(0)-4(3x^2)}{(x^3)^2}=\frac{12x^2}{x^6}=\frac{12}{x^4}$

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Question

Find $f_{xyz}$ for the function $f(x,y,z)=e^{xy}cos(2z)+xz$

Answer

For this problem, we're asked to take the derivative of a function multiple times, each time with respect to a particular variable.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant:

For the function

$f(x,y,z)=e^{xy}cos(2z)+xz$

We'll make use of the following derivative properties:

Derivative of an exponential:

Product rule:

Note that u and v may represent large functions, and not just individual variables!

Since we're looking for $f_{xyz}$ begin by taking the derivative with respect to x:

$f_x=ye^{xy}cos(2z)+z$

Now take the derivative with respect to y; be sure to utilize the product rule:

$f_{xy}=e^{xy}cos(2z)+xye^{xy}cos(2z)$

Notice that the derivative of the term when treated as a constant is zero!

Finally, take the derivative with respect to z:

$f_{xyz}=-2e^{xy}sin(2z)-2xye^{xy}sin(2z)$

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Question

Take the derivative with respect to y of the function $f(x,y,z)=x^2yz^3\cos(xy^2z)$

Answer

Note that for this problem, we're told to take the derivative with respect to one particular variable. This is known as taking a partial derivative; often it is denoted with the Greek character delta, $\delta$ , or by the subscript of the variable being considered such as or .

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: $d[\cos(u)]=-\sin(u)du$

Product rule:

Note that and may represent large functions, and not just individual variables!

For the function

$f(x,y,z)=x^2yz^3\cos(xy^2z)$

$\frac{\delta f(x,y,z)}{\delta y} = x^2z^3\cos(xy^2z)-2x^3y^2z^4\sin(xy^2z)$

$\frac{\delta f(x,y,z)}{\delta y} = x^2z^3(\cos(xy^2z)-2xy^2z\sin(xy^2z))$

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Question

Find the derivative.

Answer

Use the product rule to find this derivative.

Thus, the derivative is .

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Question

Find the derivative.

$\frac{9}{3x}$

Answer

Use the quotient rule to find the derivative.

$\frac{3x(0)-9(3)}{(3x)^2}=\frac{-27}{9x^2}$

Simplify.

The derivative is $\frac{-3}{x^2}$ .

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Question

What is the first derivative of f(x) = sin(x)ln(cos(x))?

Answer

This is a mixture of the product rule and the chain rule:

The first term of the product rule is: cos(x)ln(cos(x))

The second term will have sin(x) but will include the differentiation of the ln(cos(x)), which will require the chain rule:

sin(x) * (1/cos(x)) * (–sin(x)) = –sin2(x)/cos(x)

Combining both we get:

cos(x)ln(cos(x)) – sin2(x)/cos(x)

Now, note that none of the answers are the same as this ;however, we can make an alteration:

sin(x)/cos(x) is the same as tan(x)

Therefore, the answer is: cos(x)ln(cos(x)) – tan(x)sin(x)

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Question

What is the first derivative of f(x) = sin(cos(tan(sin(x))))

Answer

Okay, don't be overwhelmed. Take this chain rule one step at a time:

Step 1: Do the sine...

cos(cos(tan(sin(x))))

Step 2: Do the cosine . . .

–sin(tan(sin(x)))

Step 3: do the tangent . . . this is the simple chain rule, so diffentiate the argument as well

sec2(sin(x))cos(x)

Step 4: Multiply them together:

–cos(cos(tan(sin(x)))) * sec2(sin(x))cos(x) * sin(tan(sin(x)))

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Question

What is the first derivative of f(x) = sec(x2 + 4x)?

Answer

This is a simple chain rule. The derivative of the secant is secant * tangent; therefore:

f'(x) = sec(x2 + 4x) * tan(x2 + 4x) * (2x + 4)

Distribute everything to get your answer: 2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)

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Question

What is the first derivative of f(x) = cos4(x2)

Answer

Consider this as a chain rule case. Do each step:

Step 1: cos4

4cos3(x2)

Step 2: cos(x2); this can be treated like a normal case of the chain rule

–sin(x2) * 2x

Combining these, we get

–8x * sin(x2)cos3(x2)

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