How to Differentiate Exponential Functions

Differentiating General Exponential Functions

Begin with a general exponential function. Begin with a basic exponential function using a variable as the base. By calculating the derivative of the general function in this way, you can use the solution as model for a full family of similar functions. y = a x {\displaystyle y=a^{x}} y=a^{{x}}

Take the natural logarithm of both sides. You need to manipulate the function to help find a standard derivative in terms of the variable x {\displaystyle x} x. This begins by taking the natural logarithm of both sides, as follows: ln ⁡ y = ln ⁡ a x {\displaystyle \ln y=\ln a^{x}} \ln y=\ln a^{{x}}

Eliminate the exponent. Using the rules of logarithms, this equation can be simplified to eliminate the exponent. The exponent within the logarithm function can be removed as a multiple in front of the logarithm, as follows: ln ⁡ y = x ln ⁡ a {\displaystyle \ln y=x\ln a} \ln y=x\ln a

Differentiate both sides and simplify. The next step is to differentiate each side with respect to x {\displaystyle x} x. Because a {\displaystyle a} a is a constant, then ln ⁡ a {\displaystyle \ln a} \ln a is also a constant. The derivative of x {\displaystyle x} x simplifies to 1, and the term disappears. The steps are as follows: ln ⁡ y = x ln ⁡ a {\displaystyle \ln y=x\ln a} \ln y=x\ln a d d x ln ⁡ y = d d x x ln ⁡ a {\displaystyle {\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln a} {\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln a 1 y d y d x = ln ⁡ a d d x x {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=\ln a{\frac {d}{dx}}x} {\frac {1}{y}}{\frac {dy}{dx}}=\ln a{\frac {d}{dx}}x 1 y d y d x = ln ⁡ a ∗ 1 {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=\ln a*1} {\frac {1}{y}}{\frac {dy}{dx}}=\ln a*1 1 y d y d x = ln ⁡ a {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=\ln a} {\frac {1}{y}}{\frac {dy}{dx}}=\ln a

Simplify to solve for the derivative. Multiply both sides by y to isolate the derivative. Using basic steps of algebra, multiply both sides of this equation by y {\displaystyle y} y. This will isolate the derivative of y {\displaystyle y} y on the left side of the equation. Then recall that y = a x {\displaystyle y=a^{x}} y=a^{x}, so substitute that value on the right side of the equation. The steps look like this: 1 y d y d x = ln ⁡ a {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=\ln a} {\frac {1}{y}}{\frac {dy}{dx}}=\ln a d y d x = y ln ⁡ a {\displaystyle {\frac {dy}{dx}}=y\ln a} {\frac {dy}{dx}}=y\ln a d y d x = a x ln ⁡ a {\displaystyle {\frac {dy}{dx}}=a^{x}\ln a} {\frac {dy}{dx}}=a^{x}\ln a

Interpret the final result. Recalling that the original function was the exponential function y = a x {\displaystyle y=a^{x}} y=a^{x}, this solution shows that the derivative of the general exponential function is a x ln ⁡ a {\displaystyle a^{x}\ln a} a^{x}\ln a. This can be expanded for any value of a {\displaystyle a} a, as in the following examples: d d x 2 x = 2 x ln ⁡ 2 {\displaystyle {\frac {d}{dx}}2^{x}=2^{x}\ln 2} {\frac {d}{dx}}2^{x}=2^{x}\ln 2 d d x 3 x = 3 x ln ⁡ 3 {\displaystyle {\frac {d}{dx}}3^{x}=3^{x}\ln 3} {\frac {d}{dx}}3^{x}=3^{x}\ln 3 d d x 10 x = 10 x ln ⁡ 10 {\displaystyle {\frac {d}{dx}}10^{x}=10^{x}\ln 10} {\frac {d}{dx}}10^{x}=10^{x}\ln 10

Extending the Proof for the Derivative of ex

Choose the special example. The prior section showed how to differentiate the general case of an exponential function with any constant as the base. Next, select the special case where the base is the exponential constant e {\displaystyle e} e. e {\displaystyle e} e is the mathematical constant that is approximately equal to 2.718. For this derivation, select the special function y = e x {\displaystyle y=e^{x}} y=e^{x}.

Use the proof of the general exponential function derivative. Recall, from the prior section, that the derivative of a general exponential function a x {\displaystyle a^{x}} a^{x} is a x ln ⁡ a {\displaystyle a^{x}\ln a} a^{x}\ln a. Apply this result to the special function e x {\displaystyle e^{x}} e^{x} as follows: y = e x {\displaystyle y=e^{x}} y=e^{x} d y d x = d d x e x {\displaystyle {\frac {dy}{dx}}={\frac {d}{dx}}e^{x}} {\frac {dy}{dx}}={\frac {d}{dx}}e^{x} d y d x = e x ln ⁡ e {\displaystyle {\frac {dy}{dx}}=e^{x}\ln e} {\frac {dy}{dx}}=e^{x}\ln e

Simplify the result. Recall that the natural logarithm is based on the special constant e {\displaystyle e} e. Therefore, the natural logarithm of e {\displaystyle e} e is just 1. This simplifies the derivative result as follows: d y d x = e x ln ⁡ e {\displaystyle {\frac {dy}{dx}}=e^{x}\ln e} {\frac {dy}{dx}}=e^{x}\ln e d y d x = e x ∗ 1 {\displaystyle {\frac {dy}{dx}}=e^{x}*1} {\frac {dy}{dx}}=e^{x}*1 d y d x = e x {\displaystyle {\frac {dy}{dx}}=e^{x}} {\frac {dy}{dx}}=e^{x}

Interpret the final result. This proof leads to the special case that the derivative of the function e x {\displaystyle e^{x}} e^{x} is that very function itself. Thus: d d x e x = e x {\displaystyle {\frac {d}{dx}}e^{x}=e^{x}} {\frac {d}{dx}}e^{x}=e^{x}

Finding the Derivative of e with a Functional Exponent

Define your function. For this example, you will find the general derivative of functions that have e {\displaystyle e} e raised to an exponent, when the exponent itself is a function of x {\displaystyle x} x. As an example, consider the function y = e 2 x + 3 {\displaystyle y=e^{2x+3}} y=e^{{2x+3}}.

Define the variable u {\displaystyle u} u. This solution is going to involve the chain rule of derivatives. Recall that the chain rule applies when you have one function, u ( x ) {\displaystyle u(x)} u(x) nested inside another, f ( x ) {\displaystyle f(x)} f(x), as you have here. The chain rule states: d y d x = d y d u ∗ d u d x {\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dx}}} {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dx}} In summary, you will define the exponent as a separate function u ( x ) {\displaystyle u(x)} u(x). For this example, the exponent is the nested function u ( x ) {\displaystyle u(x)} u(x). Thus, for this example: y = e u {\displaystyle y=e^{u}} y=e^{u}, and u = 2 x + 3 {\displaystyle u=2x+3} u=2x+3

Apply the chain rule. The chain rule requires you to find the derivatives of both functions y {\displaystyle y} y and u {\displaystyle u} u. The resulting derivative is then the product of those two. The two separate derivatives are: d y d u = d d u e u = e u {\displaystyle {\frac {dy}{du}}={\frac {d}{du}}e^{u}=e^{u}} {\displaystyle {\frac {dy}{du}}={\frac {d}{du}}e^{u}=e^{u}}. (Remember that the derivative of e x {\displaystyle e^{x}} e^{x} is e x {\displaystyle e^{x}} e^{x}.) d u d x = d d x ( 2 x + 3 ) = 2 {\displaystyle {\frac {du}{dx}}={\frac {d}{dx}}(2x+3)=2} {\frac {du}{dx}}={\frac {d}{dx}}(2x+3)=2 After finding the two separate derivatives, combine them to find the derivative of the original function: d y d x = d y d u ∗ d u d x {\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dx}}} {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dx}} d d x e 2 x + 3 = e ( 2 x + 3 ) ∗ 2 = 2 e ( 2 x + 3 ) {\displaystyle {\frac {d}{dx}}e^{2x+3}=e^{(2x+3)}*2=2e^{(2x+3)}} {\frac {d}{dx}}e^{{2x+3}}=e^{{(2x+3)}}*2=2e^{{(2x+3)}}

Practice another example of e {\displaystyle e} e with a functional exponent. Select another example, y = e sin ⁡ x {\displaystyle y=e^{\sin x}} y=e^{{\sin x}}. Define the nested function. In this case, u = sin ⁡ x {\displaystyle u=\sin x} u=\sin x. Find the derivatives of the functions y {\displaystyle y} y and u {\displaystyle u} u. d y d u = e u {\displaystyle {\frac {dy}{du}}=e^{u}} {\frac {dy}{du}}=e^{u} d u d x = cos ⁡ x {\displaystyle {\frac {du}{dx}}=\cos x} {\frac {du}{dx}}=\cos x Combine using the chain rule: y = e sin ⁡ x {\displaystyle y=e^{\sin x}} y=e^{{\sin x}} d y d x = d y d u ∗ d u d x {\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dx}}} {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dx}} d d x e sin ⁡ x = e u ∗ cos ⁡ x = e sin ⁡ x cos ⁡ x {\displaystyle {\frac {d}{dx}}e^{\sin x}=e^{u}*\cos x=e^{\sin x}\cos x} {\frac {d}{dx}}e^{{\sin x}}=e^{u}*\cos x=e^{{\sin x}}\cos x

Finding the Derivative of xx

Define the function. For this special example, sometimes called the “power tower,” choose the function such that: y = x x {\displaystyle y=x^{x}} y=x^{x}

Find the natural logarithm of each side. As before, the solution here begins with the natural logarithm of each side of the equation: ln ⁡ y = ln ⁡ ( x x ) {\displaystyle \ln y=\ln(x^{x})} \ln y=\ln(x^{x}) ln ⁡ y = x ln ⁡ x {\displaystyle \ln y=x\ln x} \ln y=x\ln x

Take the derivative of each side of the equation. On the right side of this equation, you will need to apply the product rule of derivatives. Recall that the product rule states that if y = f ( x ) ∗ g ( x ) {\displaystyle y=f(x)*g(x)} y=f(x)*g(x), then y ′ = f ∗ g ′ + f ′ ∗ g {\displaystyle y^{\prime }=f*g^{\prime }+f^{\prime }*g} y^{{\prime }}=f*g^{{\prime }}+f^{{\prime }}*g. ln ⁡ y = x ln ⁡ x {\displaystyle \ln y=x\ln x} \ln y=x\ln x 1 y d y d x = x ∗ 1 x + 1 ∗ ln ⁡ x {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=x*{\frac {1}{x}}+1*\ln x} {\frac {1}{y}}{\frac {dy}{dx}}=x*{\frac {1}{x}}+1*\ln x 1 y d y d x = 1 + ln ⁡ x {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=1+\ln x} {\frac {1}{y}}{\frac {dy}{dx}}=1+\ln x

Multiply each side by y. Isolate the derivative term on the right by multiplying both sides of the equation by y. 1 y d y d x = 1 + ln ⁡ x {\displaystyle {\frac {1}{y}}{\frac {dy}{dx}}=1+\ln x} {\frac {1}{y}}{\frac {dy}{dx}}=1+\ln x d y d x = y ∗ ( 1 + ln ⁡ x ) {\displaystyle {\frac {dy}{dx}}=y*(1+\ln x)} {\frac {dy}{dx}}=y*(1+\ln x)

Replace the original value of y. Recall from the first step that the function is y = x x {\displaystyle y=x^{x}} y=x^{x}. Replacing this term in place of y {\displaystyle y} y is the last step to find the derivative. d y d x = y ∗ ( 1 + ln ⁡ x ) {\displaystyle {\frac {dy}{dx}}=y*(1+\ln x)} {\frac {dy}{dx}}=y*(1+\ln x) d y d x = x x ( 1 + ln ⁡ x ) {\displaystyle {\frac {dy}{dx}}=x^{x}(1+\ln x)} {\frac {dy}{dx}}=x^{x}(1+\ln x) d d x x x = x x + x x ln ⁡ x {\displaystyle {\frac {d}{dx}}x^{x}=x^{x}+x^{x}\ln x} {\frac {d}{dx}}x^{x}=x^{x}+x^{x}\ln x