$\displaystyle y = xe^{2x^2 +1}$

The function is a product of the functions $\displaystyle x$ and $\displaystyle e^{2x^2+1}$, so apply the product rule: 

$\displaystyle \frac{dy}{dx}=e^{2x^2+1}\left(\frac{d}{dx}x \right ) +x\left(\frac{d}{dx}e^{2x^2+1} \right )$

The derivative in the first term is $\displaystyle 1.$ The derivative of the second term requires the use of the chain rule. 

$\displaystyle \frac{dy}{dx}=e^{2x^2+1} +x\left(4xe^{2x^2+1} \right )$

The derivative in the second term required the use of the chain rule. First we write the derivative of $\displaystyle e^{2x^2+1}$ with respect to the function $\displaystyle 2x^2+1$, which is just $\displaystyle e^{2x^2+1}$. Then we multiply by the derivative of $\displaystyle 2x^2+1$ with respect to $\displaystyle x$, which is just $\displaystyle 4x$. 

$\displaystyle \frac{d}{dx}e^{2x^2+1}=4xe^{2x^2+1}$

$\displaystyle \frac{dy}{dx}=e^{2x^2+1} +4x^2e^{2x^2+1} = e^{2x^2+1}(1+4x^2)$

Therefore, 

 $\displaystyle \frac{dy}{dx}= e^{2x^2+1}(1+4x^2)$

Given that there are 2 terms in the numerator and only one in the denominator, one can split up the equation into 2 separate derivatives: 

$\displaystyle \frac{d}{dx}(\frac{6}{x})+\frac{d}{dx}(\frac{x^{3}}{x})$. 

Now we simplify these, and proceed to solve: 

$\displaystyle \frac{d}{dx}(6x^{-1})+\frac{d}{dx}(x^{2}) = 2x-6x^{-2}$

Because this problem contains two functions multiplied together that can't be simplified any further, it calls for the product rule, which states that

By applying this rule to the equation 

$\displaystyle x\ln(x)$ 

we get 

$\displaystyle \frac{d}{dx}(x\ln (x))= ((1)\ln (x)+(x)(\frac{1}{x}))=\ln(x)+1$

Because we are dealing with a quotient that cannot be simplified, we use the quotient rule, which states that if 

$\displaystyle h(x)=\frac{f(x)}{g(x)}$, 

$\displaystyle h{}'(x)=\frac{(g(x)f{}'(x)-f(x)g{}'(x))}{(g(x))^{2}}$.

By observing the given equation 

$\displaystyle h(x)=\frac{x}{x^{2}+1}$,

we can see that 

$\displaystyle f(x)=x$ 

and 

$\displaystyle g(x)=x^{2}+1$.

Therefore, the derivative is 

$\displaystyle \frac{(x^{2}+1)(1)-(x)(2x)}{(x^{2}+1)^{2}}=\frac{1-x^{2}}{x^{4}+2x^{2}+1}$.

Because we are differentiating a quotient that cannot be simplified, we must use the quotient rule, which states that if

$\displaystyle h(x)=\frac{f(x)}{g(x)}$,

then 

$\displaystyle h{}'(x)=\frac{g(x)f{}'(x)-f(x)g{}'(x)}{(g(x))^{2}}$.

By observing the given equation, 

$\displaystyle h(x)=\frac{x}{\ln (x)}$,

we see that in this case, 

$\displaystyle f(x)=x$ 

and 

$\displaystyle g(x)=\ln(x)$.

Given this information, the quotient rule tells us that 

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

This problem is a quotient rule inside of a chain rule. First, let's look at the chain rule:

.

Given this, we can deduce that since 

$\displaystyle h(x)=\sin(\frac{x^{2}}{x+1})$, 

$\displaystyle f(x)=\sin(x)$ 

and 

$\displaystyle g(x)=\frac{x^{2}}{x+1}$.

By plugging these into the chain rule formula, we get 

$\displaystyle (f o g(x)){}'=\cos(\frac{x^{2}}{x+1})((\frac{d}{dx}(\frac{x^{2}}{x+1})))$ 

To find the derivative of the

second term, we must use the quotient rule, which states that the the derivative

of a quotient is ((denominator)(derivative of numerator)-(numerator)(derivative

of denominator))/(denominator squared). Using this rules we find that 

$\displaystyle \frac{d}{dx}(\frac{x^{2}}{x+1})=\frac{((x+1)(2x)-(x^{2})(1))}{(x+1)^{2}}=\frac{x(x+2)}{(x+1)^{2}}$.

By plugging this back in, we find the final derivative to be 

$\displaystyle \frac{x(x+2)}{(x+1)^{2}}\cos(\frac{x^{2}}{x+1})$

The derivative of the function is equal to

$\displaystyle f'=\frac{x\sec^2(x)-\tan(x)-5}{x^2}$

and was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a$

To find the derivative of a quotient, you apply the quotient rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{f(x)}{g(x)})=\frac{{f'(x)g(x)-f(x)g'(x)}}{(g(x))^2}$

In our case, we have $\displaystyle f(x)=e^x$ and $\displaystyle g(x)=\cos{x}$

Using the function from the problem statement and taking its derivative, we get

$\displaystyle f'=\frac{e^x\cos{x}+e^x\sin{x}}{\cos^2{x}}$

To find the derivative of a quotient, you apply the quotient rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{f(x)}{g(x)})=\frac{{f'(x)g(x)-f(x)g'(x)}}{(g(x))^2}$

In our case, we have $\displaystyle f(x)=\cos{x}$ and $\displaystyle g(x)=2x^2$

Using the function from the problem statement and taking its derivative, we get

$\displaystyle f'=\frac{-2x^2\sin{x}-4x\cos{x}}{4x^4}$

To find the derivative of the function, you must apply the product rule. The product rule is as follows

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)$

In the first part of the expression, we have $\displaystyle f(x)=2x$ and $\displaystyle g(x)=\sin{x}$, and in the second part of the expression we have $\displaystyle f(x)=2$ and $\displaystyle g(x)=\cos{x}$

Using the product rule from above, we have 

$\displaystyle f'=2\sin{x}+2x\cos{x}-2\sin{x}=2x\cos{x}$