The shortest and simplest way to find the derivative of this function is to use the Chain Rule. The Chain Rule definition is \(\displaystyle [f(g(x))]' = f'(g(x))\cdot g'(x)\) . This is somewhat difficult to read and work with at first. Putting it in words helps though. What this definition states is that the derivative of "layered functions" is the derivative of the outer function times the derivative of the inner function. When I say "layered functions", I mean functions inside other functions. In this problem, we have the function, \(\displaystyle (e^{2x}+\sin(e^{x^3}))\), inside of a cubic function ,\(\displaystyle (u)^3\), where \(\displaystyle u\) is holding the place of the inner function. The outer function is the cubic, while the inner function is the\(\displaystyle (e^{2x}+\sin(e^{x^3}))\).

Applying the chain rule to this pair of layers means applying the power rule to the outer function, then multiplying it by the derivative of the inner function. Doing so gives

\(\displaystyle 3(inner function)^2\cdot(inner function)'\)

We will need to find the derivative of the inner function, \(\displaystyle (e^{2x}+\sin(e^{x^3}))\), but first we will write the expression using the actual inner function.

\(\displaystyle 3(e^{2x}+\sin(e^{x^3}))^2\cdot(e^{2x}+\sin(e^{x^3}))'\).

To find \(\displaystyle (e^{2x}+\sin(e^{x^3}))'\), we will take the derivative of the two terms inside separately.

The derivative of \(\displaystyle e^{2x}\) is \(\displaystyle e^{2x}\cdot(2)\). 

The derivative of \(\displaystyle \sin(e^{x^3})\) is another Chain Rule. We take the derivative of outer function, \(\displaystyle \sin\), to get \(\displaystyle \cos\) of the same inner function. Then we multiply it by the derivative of the inner function. The derivative of \(\displaystyle e^{x^3}\) is \(\displaystyle e^{x^3}\cdot(3x^2)\).

Putting these together we get the following for the derivative of\(\displaystyle (e^{2x}+\sin(e^{x^3}))\):

\(\displaystyle e^{2x}(2)+ \cos(e^{x^3})(e^{x^3}(3x^2))\)

Simplifying it, we get

\(\displaystyle 2e^{2x} + 3x^2e^{x^3}\cos(e^{x^3})\)

Putting this at the end of the original chain rule we have

\(\displaystyle f'(x)=3(e^{2x}+\sin(e^{x^3}))^2\cdot(2e^{2x}+ 3x^2e^{x^3}\cos(e^{x^3}))\)

This cannot be simplified, so it is the final answer

  \(\displaystyle \frac{d}{du}[sin(u)]=cos(u)\)

\(\displaystyle \frac{d}{du}[e^u]=e^u \cdot u'\)

According to the chain rule   \(\displaystyle \frac{d}{du}[f(g(u))]=f'(g(u)) \cdot g'(u)\).

Therefore, the derivative we are looking for will be 

\(\displaystyle cos(e^x) \cdot e^x \cdot 1 = e^x(cos(e^x))\)

\(\displaystyle e^{x}-cos(y)=x\)

Differentiate both sides of the equation: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(e^{x}-cos(y)=x)\)

Simplify:\(\displaystyle e^{x}-\frac{\mathrm{d} }{\mathrm{d} x}cos(y)=1\)

Use implicit differentiation to evaluate \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}cos(y)\): \(\displaystyle e^{x}-(-siny)\frac{\mathrm{d} y}{\mathrm{d} x}=1\)

Simplify:\(\displaystyle e^{x}+siny\frac{\mathrm{d} y}{\mathrm{d} x}=1\)

Subtract \(\displaystyle e^{x}\) from both sides of the equation: \(\displaystyle siny\frac{\mathrm{d} y}{\mathrm{d} x}=1-e^{x}\)

Divide both sides of the equation by siny: \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-e^{x}}{sin(y)}\)

Simplify: \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=1-e^{x}(csc(y))\)

Solution: \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=1-e^{x}(csc(y))\)

\(\displaystyle y =x^2\sin(3x)\)

Differentiate both sides and proceed with the product rule: 

\(\displaystyle \frac{dy}{dx}=\frac{d}{dx}[x^2\sin(3x)]\)                                                                                      (1)

\(\displaystyle \frac{dy}{dx}= x^2\frac{d}{dx}\sin(3x)+ \sin(3x)\frac{d}{dx}x^2\)

Evaluate the derivatives in each term. For the first term,  

\(\displaystyle =x^2\underbrace{\frac{d}{dx}\sin(3x)} + \sin(3x)\frac{d}{dx}x^2\)                                                (2)

 apply the chain rule, 

\(\displaystyle \frac{d}{dx}sin(3x)=3\cos(3x)\)

So now the first term in equation (2) can be written, 

\(\displaystyle x^2\frac{d}{dx}\sin(3x)=x^2[3\cos(3x)]=3x^2\cos(3x)\)                            (3)

The second term in equation (2) is easy, this is just the product of  \(\displaystyle \sin(3x)\)multiplied by the derivative of \(\displaystyle x^2\), 

\(\displaystyle \sin(3x)\frac{d}{dx}x^2 = 2x\sin(3x)\)                                                               (4)

Combine equations (3) and (4) to write the derivative, 

\(\displaystyle \frac{dy}{dx} =3x^2\cos(3x)+2x\sin(3x)\)

Use the product rule to find the derivative. 

\(\displaystyle \cos (x)x^2+2x\sin (x)\)

Use the power rule to find the derivative.

\(\displaystyle \frac{d}{dx}x^3=3x^2\)

\(\displaystyle \frac{d}{dx}4x^2=8x\)

\(\displaystyle \frac{d}{dx}-x=-1\)

Thus, the derivative is \(\displaystyle 3x^2+8x-1\)

Here we use the product rule: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(F(x)\cdot G(x))=F'(x)\cdot G(x)+F(x)\cdot G'(x)\)

Let \(\displaystyle F(x)=3sin(x^2)\) and \(\displaystyle G(x)=e^{2x-1}\)

Then \(\displaystyle F'(x)=3cos(x^2)\cdot2x\) (using the chain rule)

and \(\displaystyle G'(x)=e^{2x-1}\cdot2\) (using the chain rule)

Subbing these values back into our equation gives us

\(\displaystyle f'(x)=3cos(x^2)\cdot2x\cdot e^{2x-1}+3sin(x^2)\cdot e^{2x-1}\cdot2\)

Simplify by combining like-terms

\(\displaystyle f'(x)=6xcos(x^2)\cdot e^{2x-1}+6sin(x^2)\cdot e^{2x-1}\)

and pulling out a \(\displaystyle 6e^{2x-1}\) from each term gives our final answer

\(\displaystyle f'(x)=6e^{2x-1}(xcos(x^2)+sin(x^2))\)

When evaluating the derivative, pay attention to the fact that \(\displaystyle e,\pi\) are constants, (not variables) and are treated as such.

\(\displaystyle f'(x)=\pi\cos(ex)\times (ex)'=e\pi\cos(ex)\).

and hence

\(\displaystyle f'(0)=e\pi\cos(e(0))=e\pi\).

To obtain an expression for \(\displaystyle f'(x)\), we can take the derivative of \(\displaystyle f(x)\) using the sum rule.

\(\displaystyle f'(x)=0+1+2x+3x^2+4x^3\).

Substituting \(\displaystyle 2\) into this equation gives us

\(\displaystyle f'(2) =0+1+4+12+32\)

\(\displaystyle =49\).

To find \(\displaystyle f'(x)\), we will need to use the quotient rule; \(\displaystyle \frac{gf'-fg'}{g^2}\).

\(\displaystyle f(x)= \frac{x\tan^{-1}x}{\ln(x)}\). Start

\(\displaystyle f'(x)= \frac{(\ln(x))(x\tan^{-1}x)'-(x\tan^{-1}x)(\ln(x))'}{(\ln(x))^2}\). Use the quotient rule.

\(\displaystyle = \frac{(\ln(x))(x(\frac{1}{1+x^2})+1(\tan^{-1}x))-(x\tan^{-1}x)(\frac{1}{x})}{(\ln(x))^2}\). Take the derivatives inside of the quotient rule. The derivative of \(\displaystyle x\tan^{-1}x\) uses the product rule.

\(\displaystyle =\frac{\ln(x)(\frac{x}{1+x^2}+\tan^{-1}x)-\tan^{-1}x}{(\ln(x))^2}\). Simplify to match the correct answer.