Derivatives of Exponential and Logarithmic Functions: Learn It 2

Derivative of the Logarithmic Function

Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.

derivative of the natural logarithmic function

If x>0 and y=lnx, then

dydx=1x

 

More generally, let g(x) be a differentiable function. For all values of x for which g(x)>0, the derivative of h(x)=ln(g(x)) is given by

h(x)=1g(x)g(x)

Proof


If x>0 and y=lnx, then ey=x. Differentiating both sides of this equation results in the equation

eydydx=1

 

Solving for dydx yields

dydx=1ey

 

Finally, we substitute x=ey to obtain

dydx=1x

 

We may also derive this result by applying the inverse function theorem, as follows. Since y=g(x)=lnx is the inverse of f(x)=ex, by applying the inverse function theorem we have

dydx=1f(g(x))=1elnx=1x

 

Using this result and applying the chain rule to h(x)=ln(g(x)) yields

h(x)=1g(x)g(x)

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The graph of y=lnx and its derivative dydx=1x are shown in Figure 3.

Graph of the function ln x along with its derivative 1/x. The function ln x is increasing on (0, + ∞). Its derivative is decreasing but greater than 0 on (0, + ∞).
Figure 3. The function y=lnx is increasing on (0,+). Its derivative y=1x is greater than zero on (0,+).

Find the derivative of f(x)=ln(x3+3x4)

Find the derivative of f(x)=ln(x2sinx2x+1)

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y=logbx and y=bx for b>0,b1.

derivatives of general exponential and logarithmic functions

Let b>0,b1, and let g(x) be a differentiable function.

  1. If y=logbx, then
    dydx=1xlnb

    More generally, if h(x)=logb(g(x)), then for all values of x for which g(x)>0,

    h(x)=g(x)g(x)lnb

  2. If y=bx, then
    dydx=bxlnb

    More generally, if h(x)=bg(x), then

    h(x)=bg(x)g(x)lnb

Proof


If y=logbx, then by=x. It follows that ln(by)=lnx. Thus ylnb=lnx. Solving for y, we have y=lnxlnb. Differentiating and keeping in mind that lnb is a constant, we see that

dydx=1xlnb

 

The derivative from above now follows from the chain rule.

If y=bx, then lny=xlnb. Using implicit differentiation, again keeping in mind that lnb is constant, it follows that 1ydydx=lnb. Solving for dydx and substituting y=bx, we see that

dydx=ylnb=bxlnb

 

The more general derivative follows from the chain rule.

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Find the derivative of h(x)=3x3x+2

Find the slope of the line tangent to the graph of y=log2(3x+1) at x=1.