Differentiation of Exponential Functions

What is an exponential function? Exponential functions are functions that have functions in the exponents of the function. They are of a general form $f(x) = g(x)^{h(x)}$.

Then how do we take the derivative of an exponential function? When we first see an exponential function, it is often effective to express the function in logarithmic form to reduce the function to a product form: (see the wiki Properties of Logarithms)

$\ln\big(f(x)\big) = h(x)\ln\big(g(x)\big).$

Now that we have the function in a product form, we can invoke the product rule for differentiation: (see Differentiation Product Rule)

$\ln\big(f(x)\big)' = h(x)'\ln\big(g(x)\big) +h(x)\Big(\ln\big(g(x)\big)\Big)'.$

We can now invoke the differentiation rules for logarithms: (see Derivatives of Logarithmic Functions)

$\frac{f(x)'}{f(x)} = h(x)'\ln\big(g(x)\big) + \frac{h(x)g(x)'}{g(x)}.$

Solving for $f(x)$' gives

$f(x)' = f(x) h(x)'\ln\big(g(x)\big) + f(x)\frac{h(x)g(x)'}{g(x)}. \qquad (1)$

Now, often $g(x)$ may be a constant, and then we get

$f(x)' = f(x) h(x)'\ln\big(g(x)\big). \qquad (2)$

Find the derivative of $f(x) = 2\cdot3^x.$

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From (2) above, $g(x) = 3$ and $h(x) =x,$ so$f(x)' = 2\cdot3^x\ln(3).$ $_\square$

Find the derivative of $f(x) = \cos(x)\cdot e^{\ln\big(\sin(x)\big)}.$

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From (2) above,

$g(x) = e,\ \ln\big(g(x)\big) =1,\ h(x) = \ln\big(\sin(x)\big),\ h(x)' = \frac{\cos(x)}{\sin(x)}.$

Therefore, from the product rule of differentiation above,

$f(x)' = \big(\cos(x)\big)'\cdot e^{\ln\left(\sin(x)\right)} +\cos(x)\cdot \left(e^{\ln\left(\sin(x)\right)}\right)'.$

$\big($See Applying Differentiation Rules to Trigonometric Functions and rule (2) above$\big).$

We have

$\big(\cos(x)\big)' = -\sin(x), \qquad \left(e^{\ln\left(\sin(x)\right)}\right)' = e^{\ln\left(\sin(x)\right)}\cdot\frac{\cos(x)}{\sin(x)}.$

Now from (2) above, we get

$f(x)' = -\sin(x) \cdot e^{\ln\left(\sin(x)\right)} + e^{\ln\left(\sin(x)\right)} \cdot \frac{\big(\cos(x)\big)^2}{ \sin(x)}.$

Factoring and finding a common denominator, and noting $e^{\ln\left(\sin(x)\right)} = \sin(x),$ we get

$f(x)' = \big( \cos(x)\big)^2-\big(\sin(x)\big)^2.$

Simplifying, $f(x)' = \cos(2x).$ $_\square$

Find the derivative of $f(x) = \big(\sin(x)\big)^x .$

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From (1) above, $g(x) = \sin(x)$ and $h(x) =x.$ Now, since $g(x)$ is not a constant, we use derivation (1) from above, where $h(x)' =1, g(x)' = \cos(x).$ We get

$f(x)' = \big(\sin(x)\big)^x \cdot \Big(\ln\left(\sin(x)\right) + x\cdot \cot(x)\Big).\ _\square$