It measures how fast the inverse function changes with respect to its input, based on the rate of change of the original function.
The derivative of the inverse function at a point is the reciprocal of the derivative of the original function at the corresponding point.
No, the original function must be one-to-one and differentiable, and its derivative should not be zero at that point.
If the derivative is zero, the slope is flat, and the inverse function would not have a well-defined rate of change at that point.
It helps in solving problems involving logarithmic, exponential, trigonometric, and other inverse functions, especially when you need their slopes.
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Derivatives of Inverse Functions
Differentiation is one of the most important topics in JEE Main and JEE Advanced Mathematics. Most students know how to use basic rules like the power rule, product rule, and chain rule. However, one important idea is how to find the derivatives of inverse functions.
Inverse functions, particularly inverse trigonometric functions, are commonly encountered in JEE problems related to limits, continuity, and integrals. For quick and accurate problem solving, it is important to know how to derive and use their differentiation formulas.
1.0What are Inverse Functions?
If a function f(x) maps an element x to y, then the inverse function f−1(x) maps y back to x.
Example: If f(x)=ex, then its inverse isf−1(x)=ln(x)
Example: If f(x)=sinx , then its inverse is f−1(x)=sin−1x
For a function to have an inverse, it must be one-to-one and onto.
2.0Derivatives of Inverse Functions Formula
If y=f(x) is differentiable and invertible, then the derivative of its inverse is given by:
dxd[f−1(x)]=f′(f−1(x))1
This is the Derivatives of Inverse Functions Formula.
3.0Proof of Derivatives of Inverse Functions
y=f−1(x) This means:
x=f(y)
Differentiating both sides with respect to x:
1=f′(y)⋅dxdy
dxdy=f′(y)1
Since y=f−1(x)
dxd[f−1(x)]=f′(f−1(x))1
Proof complete.
4.0How to Find Derivatives of Inverse Functions?
Step-by-step process:
Identify the inverse function f−1(x)
Write y=f−1(x)⟹x=f(y)
Differentiate both sides with respect to x.
Solve for dy/dx.
Substitute y=f−1(x) back if needed.
Example: Derivative of lnx
Since ln x is the inverse of ex:ex
dxd(lnx)=dyd(ey)1y=lnx
=ey1y=lnx
=x1
Hence, dxd(lnx)=x1
5.0Derivatives of Inverse Trigonometric Functions
Inverse trigonometric functions are extremely important in JEE. Let’s derive and list them.
(i) Derivative of y=sin−1x
x=sin y
Differentiating:
1=cosy⋅dxdy
dxdy=cosy1
Since cosy=1−sin2y=1−x2
dxd(sin−1x)=1−x21
(ii) Derivative of y=cos−1x
x=cos y
Differentiating:
1=−siny⋅dxdy
dxdy=−siny1
Since siny=1−x2
dxd(cos−1x)=−1−x21
(iii) Derivative of y=tan−1x
x=tany
1=sec2y⋅dxdy
dxdy=sec2y1=1+tan2y1=1+x21
So,
dxd(tan−1x)=1+x21
(iv) Derivative of y=cot−1x
x=coty
1=−csc2y⋅dxdy
dxdy=−1+x21
(v) Derivative of y=sec−1x
x=secy
1=secytany⋅dxdy
dxd(sec−1x)=xx2−11
(vi) Derivative of y=csc−1x
x=cscy
1=−cscycoty⋅dxdy
dxdy=∣x∣x2−11
6.0Derivatives of Inverse Functions Examples (Solved)
Example 1: Find derivative of y=sin−1(1+x22x)
Solution: Let
y=sin−1(1+x22x)
This is a standard identity:
sin−1(1+x22x)=2tan−1x
So,
dxdy=dxd[2tan−1(x)]=1+x22
Example 2: If ,y=tan−1(1−x2) find dy/dx.
Solution:
y=tan−1(1−x2)
dxdy=1+(1−x2)21⋅dxd(1−x2)
=1+(1−x2)1⋅(1−x2−x)
=2−x21⋅(1−x2−x)
=(2−x2)1−x2−x
7.0Practice Questions
Differentiate y=sin−1(1−x2)
Using the formula for derivatives of inverse functions, finddxd(ln(x2+1))
Prove that dxd(tan−1x+cot−1x)=0
Differentiate y=cos−1(1+x21−x2)
Find the derivative of y=sec−1(x2+1)
Table of Contents
1.0What are Inverse Functions?
2.0Derivatives of Inverse Functions Formula
3.0Proof of Derivatives of Inverse Functions
4.0How to Find Derivatives of Inverse Functions?
4.1Example: Derivative of lnx
5.0Derivatives of Inverse Trigonometric Functions
5.1(i) Derivative of
5.2(ii) Derivative of
5.3(iii) Derivative of
5.4(iv) Derivative of
5.5(v) Derivative of
5.6(vi) Derivative of
6.0Derivatives of Inverse Functions Examples (Solved)