Examples with detailed solutions on how to find the derivative (Differentiation) of an inverse function, in calculus, are presented.
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Example 1
Find the derivative dy/dx of the inverse of function f defined by
f(x)= (1/2) x - 1
Solution to Example 1
We present two methods to answer the above question
Method 1
The first method consists in finding the inverse of function f and differentiate it. To find the inverse of f we first write it as an equation
y = (1/2) x - 1
Solve for x.
x = 2y + 2.
Change y to x and x to y.
y = 2x + 2.
The above gives the inverse function of f. Let us find the derivative
dy / dx = 2
Method 2
The second method starts with one of the most important properties of inverse functions.
f(f -1(x)) = x
Let y = f -1(x) so that.
f(y) = x.
Differentiate both sides using chain rule to the left side.
(dy/dx)(df/dy) = 1.
Solve for dy/dx
dy / dx = 1 / (df / dy)
f is defined by
f(x)= (1/2) x - 1
so that df / dy = 1/2
Substitute df / dy by 1/2 in dy / dx = 1 / (df / dy) to obtain
dy / dx = 1 / (1/2) = 2
The first method can be used only if we can find explicitly the inverse function.
Example 2
Find the derivative dy / dx where y = arcsin x.
Solution to Example 2
arcsin x is the inverse function of sin x and
sin(arcsin(x)) = x
y = arcsin x so that
sin y = x
Differentiate both sides of the above equation, with respect to x, using the chain rule on the left side.
dy/dx cos y = 1
Solve for dy/dx.
dy/dx = 1 / cos y
= 1 / cos ( arcsin x)
= 1 / √(1 - sin 2(arcsin x))
= 1 / √ (1 - x 2)
Exercises
Find the derivative of the inverse of each function.
1) f(x) = 3x - 4
2) g(x) = arccos x
3) h(x) = arctan x
solutions to the above exercises
1) f -1 ' (x) = 1 / 3
2) g -1 ' (x) = -1 / √(1 - x 2)
3) h -1 ' (x) = 1 / (1 + x 2)
