Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

\(\displaystyle x^y=yx\)

\(\displaystyle \ln(x^y)=\ln(xy)\)

\(\displaystyle y\ln(x)=\ln(x)+\ln(y)\)

\(\displaystyle y = \frac{\ln(y)}{\ln(x)}+1\)

\(\displaystyle y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}\). Quotient Rule + implicit differentiation.

\(\displaystyle y' = \frac{y'}{y\ln(x)}-\frac{\ln(y)}{x\ln(x)^2}\)

\(\displaystyle y' -\frac{1}{y\ln(x)}y' = -\frac{\ln(y)}{x\ln(x)^2}\)

\(\displaystyle (\frac{1}{y\ln(x)}-1)y' = \frac{\ln(y)}{x\ln(x)^2}\)

\(\displaystyle (\frac{1-y\ln(x)}{y\ln(x)})y' = \frac{\ln(y)}{x\ln(x)^2}\)

\(\displaystyle y' = \frac{\ln(y)}{x\ln(x)^2}(\frac{y\ln(x)}{1-y\ln(x)})\)

\(\displaystyle y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}\)

The correct answer is \(\displaystyle \frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}\).

Using the Quotient Rule and the fact \(\displaystyle \frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}\), we have

\(\displaystyle y = \frac{\tan^{-1}(x)}{x}\)

\(\displaystyle y' = \frac{x\frac{1}{1+x^2}-\tan^{-1}(x)}{x^2}\)

\(\displaystyle y' = \frac{\frac{x}{1+x^2}-\tan^{-1}(x)}{x^2}\)

\(\displaystyle y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)}{x^2}\)

\(\displaystyle y' = \frac{x}{x^2(1+x^2)}-\frac{\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}\)

\(\displaystyle y' = \frac{x-\tan^{-1}(x)(1+x^2)}{x^2(1+x^2)}\).

First, factor out \(\displaystyle e^x\): \(\displaystyle e^x (2x^2 -x )\). Now we can differentiate using the product rule, \(\displaystyle f(x)g(x) '= f'(x)g(x) + f(x)g'(x)\). 

Here, \(\displaystyle f(x) = e^x\) so \(\displaystyle f'(x) = e^x\). \(\displaystyle g(x) = 2x^2 - x\) so \(\displaystyle g'(x) = 4x - 1\). 

The answer is \(\displaystyle e^x (2x^2 -x) + e^x (4x - 1 ) = e^x [ 2x^2 - x + 4x - 1 ] = e^x (2x^2 + 3x -1)\)

According to the product rule, \(\displaystyle f(x)g(x) ' = f'(x)g(x) + f(x)g'(x)\). Here \(\displaystyle f(x) = 2 \sin x\) so \(\displaystyle f'(x) = 2 \cos x\). \(\displaystyle g(x) = \cos x\) so \(\displaystyle g'(x) = -\sin x\). 

The derivative is \(\displaystyle 2 \cos x \cdot \cos x + 2 \sin x \cdot - \sin x = 2 \cos ^2 x - 2 \sin ^2 x\)

Factoring out the 2 gives \(\displaystyle 2(\cos ^2 x - \sin ^2 x )\). Remembering the double angle trigonometric identity finally gives \(\displaystyle 2 (\cos (2x ))\)

First, we need to find \(\displaystyle f'(t)\). We can do that by using the quotient rule.

\(\displaystyle f'(t) = \frac{(\ln(t))(1)-(t)(\frac{1}{t})}{(\ln(t))^2}\).

Plugging \(\displaystyle e^2\) in for \(\displaystyle t\) and simplifying, we get

\(\displaystyle f'(e^2) = \frac{(\ln(e^2))(1)-(e^2)(\frac{1}{e^2})}{(\ln(e^2))^2} = \frac{2-1}{2^2} = \frac{1}{4}\).

The derivative of the function is equal to

\(\displaystyle f'=\frac{2x^3\cos(x)+x^4\sin(x)}{x^2\cos^2(x)}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

When taking the derivative of the sum, we simply take the derivative of each component. 

The derivative of the function is

\(\displaystyle f'=3x^2+2x+2\alpha^2\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\)

Compute the first derivative of the following function.

\(\displaystyle f(x)=sin(x)cox(x)+x^2sin(x)\)

To solve this problem, we need to apply the product rule:

\(\displaystyle (f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\)

So, we need to apply this rule to each of the terms in our function. Let's start with the first term

\(\displaystyle (sin(x)cos(x))'=cos(x)cos(x)+sin(x)(-sin(x))=cos^2(x)-sin^2(x)\)

Next, let's tackle the second part

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

Now, combine the two to get:

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

Use the product rule: \(\displaystyle (fg)' = f'g + fg'\) 

where \(\displaystyle f(x) = x^2\) and \(\displaystyle g(x) = \cos(3x)\).

By the power rule, \(\displaystyle f'(x)=2x\). 

By the chain rule, \(\displaystyle g'(x) = -\sin(3x)\cdot3\).

Therefore, the derivative of the entire function is:

\(\displaystyle y'=(2x)(\cos(3x))+(x^2)(-\sin(3x)\cdot 3)\)

\(\displaystyle y'=2x\cos(3x)-3x^2\sin(3x)\).

Find the second derivative of g(x)

\(\displaystyle g(x)=3x^4ln(x)\)

To find this derivative, we need to use the product rule:

\(\displaystyle (f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\)

So, let's begin:

\(\displaystyle g(x)=3x^4ln(x)\)

\(\displaystyle g'(x)=12x^3ln(x)+3x^4(\frac{1}{x})=12x^3ln(x)+3x^3\)

So, we are closer, but we need to derive again to get the 2nd derivative

\(\displaystyle g'(x)=12x^3ln(x)+3x^3\)

\(\displaystyle g''(x)=36x^2ln(x)+12x^3(\frac{1}{x})+9x^2=36x^2ln(x)+21x^2\)

So, our answer is:

\(\displaystyle 36x^2ln(x)+21x^2\)