Recognize the derivatives of the standard inverse trigonometric functions
We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.
Example: Derivative of the Inverse Sine Function
Use the inverse function theorem to find the derivative of g(x)=sin−1x.
Show Solution
Since for x in the interval [−π2,π2],f(x)=sinx is the inverse of g(x)=sin−1x, begin by finding f′(x). Since
f′(x)=cosx and f′(g(x))=cos(sin−1x)=√1−x2,
we see that
g′(x)=ddx(sin−1x)=1f′(g(x))=1√1−x2
Analysis
To see that cos(sin−1x)=√1−x2, consider the following argument. Set sin−1x=θ. In this case, sinθ=x where −π2≤θ≤π2. We begin by considering the case where 0<θ<π2. Since θ is an acute angle, we may construct a right triangle having acute angle θ, a hypotenuse of length 1, and the side opposite angle θ having length x. From the Pythagorean theorem, the side adjacent to angle θ has length √1−x2. This triangle is shown in Figure 2. Using the triangle, we see that cos(sin−1x)=cosθ=√1−x2.
Figure 2. Using a right triangle having acute angle θ, a hypotenuse of length 1, and the side opposite angle θ having length x, we can see that cos(sin−1x)=cosθ=√1−x2.
In the case where −π2<θ<0, we make the observation that 0<−θ<π2 and hence cos(sin−1x)=cosθ=cos(−θ)=√1−x2.
Now if θ=π2 or θ=−π2,x=1 or x=−1, and since in either case cosθ=0 and √1−x2=0, we have
cos(sin−1x)=cosθ=√1−x2
Consequently, in all cases, cos(sin−1x)=√1−x2.
Example: Applying the Chain Rule to the Inverse Sine Function
Apply the chain rule to the formula derived in Example: Applying the Inverse Function Theorem
to find the derivative of h(x)=sin−1(g(x)) and use this result to find the derivative of h(x)=sin−1(2x3).
Show Solution
Applying the chain rule to h(x)=sin−1(g(x)), we have
h′(x)=1√1−(g(x))2g′(x).
Now let g(x)=2x3, so g′(x)=6x2. Substituting into the previous result, we obtain
h′(x)=1√1−4x6⋅6x2=6x2√1−4x6
Watch the following video to see the worked solution to Example: Applying the Chain Rule to the Inverse Sine Function.
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Use the inverse function theorem to find the derivative of g(x)=tan−1x.
Hint
The inverse of g(x) is f(x)=tanx. Use (Figure) as a guide.
Show Solution
g′(x)=11+x2
The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.
Example: Applying Differentiation Formulas to an Inverse Tangent Function
Find the derivative of f(x)=tan−1(x2)
Show Solution
Let g(x)=x2, so g′(x)=2x. Substituting into (Figure), we obtain
f′(x)=11+(x2)2⋅(2x)
Simplifying, we have
f′(x)=2x1+x4
Example: Applying Differentiation Formulas to an Inverse Sine Function
Find the derivative of h(x)=x2sin−1x
Show Solution
By applying the product rule, we have
h′(x)=2xsin−1x+1√1−x2⋅x2.
Watch the following video to see the worked solution to Example: Applying Differentiation Formulas to an Inverse Sine Function.
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.