# Applying Differentiation Rules To Exponential Functions

We have learned that the derivative of an exponential function is given by the formulas:

\[\begin{array} &&y=e^x &\Rightarrow y'=e^x\\ &y=e^{f(x)} &\Rightarrow y'=e^{f(x)}f'(x)\\ &y=a^x &\Rightarrow y'=a^x(\ln a)\\ &y=a^{f(x)} &\Rightarrow y'=a^{f(x)}(\ln a)\cdot f'(x), \end{array}\] where we denote \(\frac{dy}{dx}\) as \(y'\) and \(\frac{df(x)}{dx}\) as \(f'(x),\) and \(a\) is a random base such that \(a>0\) and \(a\ne1.\) When differentiating complex exponential functions, just stick to the formulas above along with the differentiation rules that we have learned earlier. To remind, here is a list of some differentiation rules: \[\begin{array} &&\bullet \ y=cf(x) &\Rightarrow y'=cf'(x)\\ &\bullet \ y=f(x)+g(x) &\Rightarrow y'=f'(x)+g'(x)\\ &\bullet \ y=f(x)g(x) &\Rightarrow y'=f'(x)g(x)+f(x)g'(x)\\ &\bullet \ y=\frac{f(x)}{g(x)} &\Rightarrow y'=\frac{g(x)f'(x)-f(x)g'(x)}{\left(g(x)\right)^2}\\ &\bullet \ y=g\left(f(x)\right) &\Rightarrow y'=g'\left(f(x)\right)\cdot f'(x). \end{array}\] Another rule that might be handy is one that we have learned when studying logarithms in algebra. Recall that:

\[a^{\log_b c}=c^{\log_b a}.\]

Now let's try differentiating complex exponential functions using these rules.

## Example #1. What is the derivative of \(y=3^{x^3+1}?\)

If \(y=a^{f(x)},\) then \(y'=a^{f(x)}(\ln a)\cdot f'(x).\) Since \(y=3^{x^3+1}\) in this problem, we have \[\begin{align} y'&=3^{x^3+1} (\ln 3) \cdot \left(x^3+1\right)'\\ &=3^{x^3+1} (\ln 3) \cdot 3x^2\\ &=3^{x^3+2}x^2 \cdot \ln 3. \ _\square \end{align}\]

## Example #2. What is the derivative of \(y=5^{\sin x}?\)

If \(y=a^{f(x)},\) then \(y'=a^{f(x)}(\ln a)\cdot f'(x).\) Since \(y=5^{\sin x}\) in this problem, we have \[\begin{align} y'&=5^{\sin x}(\ln 5) \cdot (\sin x)'\\ &=5^{\sin x}(\ln 5) \cdot (\cos x)\\ &=5^{\sin x} \cos x \cdot \ln 5. \ _\square \end{align}\]

## Example #3. What is the derivative of \(f(x)=5^{\cos3x }?\)

This problem can be solved using the same method as example #2. However, here is another way to look at it.

Converting the base to \(e\) by using the formula \(e^{\ln a^x} =a^x\) gives

\[\begin{align} f(x)&=5^{\cos3x }=e^{\ln5^{\cos3x}}=e^{\cos3x\cdot\ln5}\\ \Rightarrow f'(x)&=-3\sin3x\cdot\ln5\cdot e^{\cos3x\cdot\ln5}.\ _\square \end{align}\]

## Example #4. What is the derivative of \(y=xe^{\cos x}?\)

We have \[\begin{align} y'&=(x)'e^{\cos x}+x\left(e^{\cos x}\right)'\\ &=1\cdot e^{\cos x}+x \cdot e^{\cos x}\cdot(\cos x)'\\ &= e^{\cos x}+x\cdot e^{\cos x}\cdot(-\sin x)\\ &= e^{\cos x}(1-x\sin x).\ _\square \end{align}\]

## Example #5. What is the derivative of \(y=e^{3 x}\sin 2x?\)

We have \[\begin{align} y'&=\left(e^{3 x}\right)'\sin 2x+e^{3 x}(\sin 2x)'\\ &=3e^{3 x}\sin 2x+e^{3 x}\cdot 2\cos 2x\\ &=e^{3x}(3\sin 2x+2\cos 2x).\ _\square \end{align}\]

## Example #6. What is the derivative of \(\displaystyle{y=\frac{e^x-e^{-x}}{e^x+e^{-x}}?}\)

We have \[\begin{align} y'&=\frac{\left(e^x-e^{-x}\right)'\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x+e^{-x}\right)'}{\left(e^x+e^{-x}\right)^2}\\ &=\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}\\ &=\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x+e^{-x}\right)^2}\\ &=\frac{4}{\left(e^x+e^{-x}\right)^2}.\ _\square \end{align}\]

**Cite as:**Applying Differentiation Rules To Exponential Functions.

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