To take the derivative, you must first take the derivative of the outside function, which is sine.  However, the \(\displaystyle x^2\), or the angle of the function, remains the same until we take its derivative later.  The derivative of sinx is cosx, so you the first part of \(\displaystyle f'(x)\) will be \(\displaystyle cos(x^2)\).  Next, take the derivative of the inside function, \(\displaystyle x^2\).  Its derivative is \(\displaystyle 2x\), so by the chain rule, we multiply the derivatives of the inside and outside functions together to get \(\displaystyle f'(x)=2xcos(x^2)\).

By the chain rule, we must first take the derivative of the outside function by bringing the power down front and reducing the power by one.  When we do this, we do not change the function that is in the parentheses, or the inside function.  That means that the first part of \(\displaystyle f'(x)\) will be \(\displaystyle 2(4x^3-9x^2+4x-8)\).  Next, we must take the derivative of the inside function.  Its derivative is \(\displaystyle (12x^2-18x+4)\).  The chain rule says we must multiply the derivative of the outside function by the derivative of the inside function, so the final answer is \(\displaystyle f'(x)=2(4x^3-9x^2+4x-8)*(12x^2-18x+4)\).

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

To differentiate the equation above, start by applying the derivative operation to both sides, 

\(\displaystyle \frac{d}{dx}(xy^2)=\frac{d}{dx}[y\sin(x)]+\underset{=0}{ \underbrace{\frac{d}{dx}(1)}}\) 

Both sides will require the product rule to differentiate, 

\(\displaystyle y^2\frac{d}{dx}(x) +x\frac{d}{dx}(y^2) =\sin(x)\frac{d}{dx}(y)+y\frac{d}{dx}\sin(x)\)

\(\displaystyle y^2\underset{=1}{ \underbrace{\frac{d}{dx}(x)}} +x\left(2y\frac{dy}{dx} \right ) =\sin(x)\frac{dy}{dx}+y\cos(x)\)

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 Common Mistake 

A common mistake in the previous step would be to conclude that \(\displaystyle \frac{d}{dx}(y^2)=2y\)instead of \(\displaystyle 2y\frac{dy}{dx}\) . The former is not correct; if we were looking for the derivative with respect to \(\displaystyle y\), then \(\displaystyle \frac{d}{dy}(y^2)\) would in fact be \(\displaystyle 2y\). But we are not differentiating with respect to \(\displaystyle y\), we're looking for the derivative with respect to \(\displaystyle x\). 

We are assuming that \(\displaystyle y\) is a function of \(\displaystyle x\), so we must apply the chain rule by differentiating with respect to \(\displaystyle y\) and multiplying by the derivative of \(\displaystyle y\) with respect to \(\displaystyle x\) to obtain \(\displaystyle \frac{d}{dx}(y^2)=2y\frac{dy}{dx}\). 

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 \(\displaystyle y^2+2xy\frac{dy}{dx} =\sin(x)\frac{dy}{dx}+y\cos(x)\)

Collect terms with a derivative onto one side of the equation, factor out the derivative, and divide out  to solve for the derivative \(\displaystyle \frac{dy}{dx}\), 

\(\displaystyle 2xy\frac{dy}{dx} -\sin(x)\frac{dy}{dx}=y\cos(x)-y^2\)

\(\displaystyle \frac{dy}{dx}[2xy -\sin(x)]=y\cos(x)-y^2\)

\(\displaystyle \frac{dy}{dx}=\frac{y\cos(x)-y^2}{2xy -\sin(x)}=\frac{y(\cos x-y)}{2xy -\sin (x)}\)

Therefore, 

 \(\displaystyle \frac{dy}{dx}=\frac{y(\cos x-y)}{2xy -\sin x}\)

\(\displaystyle y = \sin(\sin x)\)                          (1)

An easier way to think about this:

Because \(\displaystyle y\) is a function of a function, we must apply the chain rule. This can be confusing at times especially for function like equation (1). The differentiation is easier to follow if you use a substitution for the inner function, 

Let,

 \(\displaystyle u = \sin(x)\)                              (2)

So now equation (1) is simply, 

\(\displaystyle y =\sin (u)\)                               (3)

Note that \(\displaystyle u\) is a function of \(\displaystyle x\). We must apply the chain rule to find  \(\displaystyle \frac{dy}{dx}\), 

\(\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\)                            (4)

To find the derivatives on the right side of equation (4), differentiate equation (3) with respect to \(\displaystyle u\), then Differentiate equation (2) with respect to \(\displaystyle x\). 

                               \(\displaystyle \frac{dy}{du}=\cos(u)\)                   \(\displaystyle \frac{du}{dx}=\cos(x)\)        

Substitute into equation (4),  

\(\displaystyle \frac{dy}{dx}=\cos(u)\cos(x)\)                  (5)

Now use \(\displaystyle u = \sin(x)\) to write equation (5) in terms of \(\displaystyle x\) alone: 

\(\displaystyle \frac{dy}{dx} =\cos(\sin(x))\times\cos(x)\)

Here we use the chain rule: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}ln(g(x))=\frac{1}{g(x)}\cdot g'(x)\)

Let \(\displaystyle g(x)=x^3+x^2\)

Then \(\displaystyle g'(x)=3x^2+2x\)

And \(\displaystyle f'(x)=\frac{1}{x^3+x^2}\cdot (3x^2+2x)=\frac{x(3x+2)}{x(x^2+x)}=\frac{3x+2}{x^2+x}\)

Using the chain rule, we have

\(\displaystyle f'(x) = \frac{1}{e^{2x^2}}\times (e^{2x^2})'\)

\(\displaystyle = \frac{1}{e^{2x^2}}\times e^{2x^2} \times 4x = 4x\).

Hence, \(\displaystyle f'(4) = 4(4)=16\).

Notice that we could have also simplified \(\displaystyle f(x)\) first by cancelling the natural log and the exponential function leaving us with just \(\displaystyle f(x) =2x^2\), thereby avoiding the chain rule altogether.

First, differentiate the outside of the parenthesis, keeping what is inside the same.

You should get  \(\displaystyle 2(3x+1)\).

Next, differentiate the inside of the parenthesis. 

You should get \(\displaystyle 3\).

Multiply these two to get the final derivative \(\displaystyle y'=6(3x+1)\).

Use chain rule to solve this. First, take the derivative of what is outside of the parenthesis.

You should get \(\displaystyle \frac{1}{cos(x)}\).

Next, take the derivative of what is inside the parenthesis. 

You should get \(\displaystyle -sin(x)\).

Multiplying these two together gives \(\displaystyle -tan(x)\).

This is a chain rule derivative.  We must first start by taking the derivative of the outermost function.  Here, that is a function raised to the fifth power.  We need to take that derivative (using the the power rule).  Then, we multiply by the derivative of the innermost function:

\(\displaystyle f'(x)=5(1+3x^2)^{5-1}*6x=30x(1+3x^2)^4.\)

This is a chain rule derivative.  We must first differentiate the natural log function, leaving the inner function as is. Recall:

\(\displaystyle ln(x)'=\frac{1}{x}\)

Now, we must replace this with our function, and multiply that by the derivative of the inner function:

\(\displaystyle ln(3x)'=\frac{1}{3x}(3)=\frac{3}{3x}=\frac{1}{x}.\)