We simply differentiate both sides of the equation

$\displaystyle sin(x^2)+cos(y^2)=1$

$\displaystyle 2xcos(x^2)dx-2ysin(y^2)dy=0$ (Don't forget the chain rule)

Now we solve for $\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}$

$\displaystyle xcos(x^2)dx=ysin(y^2)dy$

$\displaystyle xcos(x^2)=\frac{ysin(y^2)dy}{dx}$

$\displaystyle \frac{xcos(x^2)}{ysin(y^2)}=\frac{\mathrm{d}y }{\mathrm{d} x}$

Implicit differentiation is very similar to normal differentiation, but every time we take a derivative with respect t $\displaystyle y$, we need to multiply the result by $\displaystyle \frac{dy}{dx}.$  We also differentiate the entire equation from left to right, including any numbers. Then, we solve for that $\displaystyle \frac{dy}{dx}$ for our final answer.

$\displaystyle x^2+3y^2=9$

$\displaystyle 2x+6y\frac{dy}{dx}=0$

$\displaystyle \frac{dy}{dx}=\frac{-2x}{6y}=\frac{-x}{3y}.$

Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to $\displaystyle y$, we need to implicitly write $\displaystyle \frac{dy}{dx}$ after it.  Hence, the name of this method.  Then, we solve for $\displaystyle \frac{dy}{dx}.$

$\displaystyle (y^2=4sinx+3x^2)'=2y\frac{dy}{dx}=4cosx+6x.$

$\displaystyle \frac{dy}{dx}=\frac{4cosx+6x}{2y}=\frac{2cosx+3x}{y}.$

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as $\displaystyle y'$.

$\displaystyle y'=\frac{\mathrm{d} }{\mathrm{d} x}(xy^2)+\frac{\mathrm{d} }{\mathrm{d} x}(2\tan{y})$

$\displaystyle y'=y^2+y'(2xy)+2y'\sec^2(y)$

Bringing the terms with $\displaystyle y'$ to one side and factoring it out, we get

$\displaystyle y'(1-2xy-2\sec^2y)=y^2$

$\displaystyle y'=\frac{y^2}{1-2xy-2\sec^2{y}}$

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as $\displaystyle y'$.

$\displaystyle y'=\frac{\mathrm{d} }{\mathrm{d} x}(x^3y)+\frac{\mathrm{d} }{\mathrm{d} x}(y^2)$

$\displaystyle y'=y'x^3+3x^2y+2yy'$

Bringing the terms with $\displaystyle y'$ to one side and factoring it out, we get

$\displaystyle y'(1-x^3-2y)=3x^2y$

$\displaystyle y'=\frac{3x^2y}{1-x^3-2y}$

To find $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$, we must take the derivative of both sides of the equation with respect to x. When we do this, we get

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}+2x+e^xy^2+2ye^x\frac{\mathrm{d}y }{\mathrm{d} x}=4y\frac{\mathrm{d} y}{\mathrm{d} x}$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

Note that for every derivative of a function of y with respect to x, the chain rule was used, which accounts for $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$ appearing.

Algebraic simplification gets us our final answer,

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2x-e^xy^2}{1+2ye^x-4y}$

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

$\displaystyle -\sin(xy)(y+x\frac{\mathrm{d} y}{\mathrm{d} x}) =2$

The following derivative rules were used:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the chain rule was used for both the cosine function (which contains an inner product of two functions), and for the derivative of y, whose derivative with respect to x we want to solve for. 

Solving, we get

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{2+y\sin(xy)}{x\sin(xy)}$

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

$\displaystyle y^2 \frac{\mathrm{d} y}{\mathrm{d} x}+3x^2=2ye^{2x}+e^{2x}\frac{\mathrm{d} y}{\mathrm{d} x}$

The following rules were used:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the chain rule was used everywhere we took the derivative of a function containing y, as well as in the exponential function. The product rule was used because both x and y are both functions of x.

Using algebra to solve, we get

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

To find the derivative of $\displaystyle \alpha$ with respect to x, we must take the differentiate both sides of the equation with respect to x:

$\displaystyle 2\alpha x^2 \frac{\mathrm{d} \alpha}{\mathrm{d} x}+2\alpha^2x+2x \frac{\mathrm{d} \alpha}{\mathrm{d} x}+2x=\frac{\mathrm{d} \alpha}{\mathrm{d} x}$

The following derivative rules were used:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the chain rule was used for the derivative of any function containing $\displaystyle \alpha$, whose derivative with respect to x we want to solve for. 

Solving, we get

$\displaystyle \frac{\mathrm{d} \alpha}{\mathrm{d} x}=\frac{2\alpha^2 x+2 \alpha}{1-2\alpha x^2 -2x}$

To solve this using implicit differentiation, we must always treat y as a function of x, and therefore when we differentiate y with respect to x, we denote it as $\displaystyle y'$

Step by step, we get the following:

$\displaystyle y'=\cos(xy)(y+xy')+\sec^2(y^2)(2yy')$

This resulted from the product rule and chain rule

The next steps are:

$\displaystyle y'=y\cos(xy)+xy'\cos(xy)+2yy'\sec^2(y^2)$

$\displaystyle y'-xy'\cos(xy)-2yy'\sec^2(y^2)=y\cos(xy)$

$\displaystyle y'(1-x\cos(xy)-2y\sec^2(y^2))=y\cos(xy)$

$\displaystyle y'=\frac{y\cos(xy)}{1-x\cos(xy)-2y\sec^2(y^2)}$