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)
■
The graph of y=lnx and its derivative dydx=1x are shown in Figure 3.
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+3x−4)
Use the derivative of a natural logarithm directly.
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.
f(x)=ln(x2sinx2x+1)=2lnx+ln(sinx)−ln(2x+1)Apply properties of logarithms.f′(x)=2x+cosxsinx−22x+1Apply sum rule andh′(x)=1g(x)g′(x).=2x+cotx−22x+1Simplify using the quotient identity for cotangent.
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,b≠1.
derivatives of general exponential and logarithmic functions
Let b>0,b≠1, and let g(x) be a differentiable function.
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
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.
■
Find the derivative of h(x)=3x3x+2
Use the quotient rule and the derivative from above.
h′(x)=3xln3(3x+2)−3xln3(3x)(3x+2)2Apply the quotient rule.=2⋅3xln3(3x+2)2Simplify.
Find the slope of the line tangent to the graph of y=log2(3x+1) at x=1.
To find the slope, we must evaluate dydx at x=1. Using the derivative above, we see that
dydx=3ln2(3x+1)
By evaluating the derivative at x=1, we see that the tangent line has slope