Since both the numerator and denominator contain a variable, we must use the quotient rule.

\(\displaystyle f'(x)=\frac{(2^{x}){\color{Green} \frac{\mathrm{d} }{\mathrm{d} x}(5^{x^{2}})}-(5^{x^{2}}){\color{Blue} \frac{\mathrm{d} }{\mathrm{d} x}(2^{x})}}{(2^{x})^{2}}\)

Remember that the derivative of a constant raised to a variable power follows the pattern, \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} u}(a^{u})=a^{u}\ln\left | a \right | \mathrm{d} u\), where a is a constant and u is a function of x.

Applying this we get,

\(\displaystyle f'(x)= \frac{(2^{x}){\color{Green} (5^{x^{2}}(\ln5)(2x))}-(5^{x^{2}}){\color{Blue} (2^{x}(\ln2)(1))}}{(2^{x})^{2}}\)

Now we simplify by factoring the greatest common factor out of the numerator.

\(\displaystyle f'(x)=\frac{2^{x}\cdot 5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{(2^{x})^{2}}\)

Then we can cancel a single  \(\displaystyle 2^{x}\) from the numerator and denominator.

\(\displaystyle f'(x)=\frac{5^{x^{2}}(2x\cdot \ln(5)-\ln(2))}{2^{x}}\)

This is the correct answer.

Using the Quotient rule for derivatives, we know that the derivative will equal \(\displaystyle \frac{(1-e^t)(1)-t(-e^t)}{(1-e^t)^2}\).

When you simplify this, you get \(\displaystyle \frac{1-e^t+te^t}{1-2e^t+e^{2t}}\).

Using the Quotient rule for derivatives, we know that the derivative will equal :\(\displaystyle \frac{y(e^y)-e^y(1)}{y^2}\).

If you simplify this, you will get:

\(\displaystyle \frac{(y-1)e^y}{y^2}\)

Rewriting the original equation gives us \(\displaystyle f(x)=2(x^{-1})+\frac{1}{4}x^{-3}-6(x^{-5})\).

Then, one can just use the power rule to get: \(\displaystyle -2x^{-2}-\frac{3}{12}x^{-4}-30x^{-6}\).

Simplifying this gives us: \(\displaystyle -2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}\).

Step 1: Recall any rules that are used in derivatives...

• Power Rule
• Quotient Rule
• Product Rule
• Chain Rule

Step 2: Look at the choices in the question and compare the ones listed in Step 1.

We see Power Rule, Chain Rule, and Product Rule.

There is no such rule called the Addition rule, so this is the incorrect answer.

\(\displaystyle f(x)=h(x-g(x))+g(x)\)

This is a conceptual problem. First notice that the function \(\displaystyle f(x)\) is the sum of two functions \(\displaystyle h(x-g(x))\) and \(\displaystyle g(x)\). We must differentiate each term. 

To differentiate the first term, notice that \(\displaystyle h(x-g(x))\) is a function of the function \(\displaystyle x-g(x)\), so we must use the chain rule to differentiate with respect to \(\displaystyle x\). We cannot conclude that \(\displaystyle h'(x-g(x))=2x\). 

We were given \(\displaystyle h'(x)=2x\). What this equation is telling us is that the function \(\displaystyle h(x)\)  will have a derivative that is 2 times whatever is inside the parenthesis, with respect to the whatever is inside the parenthesis. If we wish to use \(\displaystyle h'(x)=2x\) to differentiate the composite  \(\displaystyle h(x-g(x))\) we can start with the derivative with respect to  \(\displaystyle x-g(x)\), which will be \(\displaystyle 2(x-g(x))\).

Now if we want to differentiate with respect to \(\displaystyle x\), we take the derivative with respect to \(\displaystyle x-g(x)\) and multiply by the derivative of \(\displaystyle x-g(x)\) with respect to \(\displaystyle x\)  (the chain rule). 

\(\displaystyle h'(x-g(x))=2(x-g(x))\times(1-g'(x))\)

We were given \(\displaystyle g'(x)=4\)

  \(\displaystyle h'(x-g(x))=2(x-g(x))\times(-3)\)

\(\displaystyle h'(x-g(x))= 6x-6g(x)\)

Now we can write the derivative of \(\displaystyle f(x)\), 

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

\(\displaystyle f'(x)=6x-6g(x)+4\)

Find the equation of the tangent line to the curve corresponding to \(\displaystyle x = 2\)

\(\displaystyle f(x)=\ln(x)+3x^2\)

The first step is to compute the derivative for the function and then evaluate the derivative at \(\displaystyle x = 2\)

\(\displaystyle f'(x)=\frac{1}{x}+6x\: \: \: \:\: \: \:\Rightarrow \: \: \: \: \: \: \: \:f'(2)=\frac{1}{2}+6(2)=\frac{25}{2}\)

Therefore \(\displaystyle \frac{25}{2}\) will be the slope of the tangent line at \(\displaystyle x = 2\). Write an equation for the line,  

\(\displaystyle y = mx+b\)

\(\displaystyle y = \frac{25}{2}x+b\)

Now we need to find the y-intercept. Use the original function to find \(\displaystyle y\) when \(\displaystyle x=2\). 

  \(\displaystyle f(2)=\ln(2)+3(2^2)=\ln(2)+12\)

This gives us the point at which the tangent line meets the curve, 

  \(\displaystyle (2\: \: ,\: \, \ln(2)+12)\)

Now use this point to find the y-intercept, 

 \(\displaystyle b = y-\frac{25}{2}x\)

 \(\displaystyle b = \ln(2)+12-\frac{25}{2}(2)\)

\(\displaystyle b=\ln(2)-13\). 

The equation of the line is therefore, 

\(\displaystyle y = \frac{25}{2}x+\ln(2) -13\)