Concept understanding — Second Derivative Inverse Cosine
Second Derivative of Inverse Cosine
What we are after
The second derivative of a function is just the derivative of its first derivative — it measures how the slope itself is changing. Here we apply that idea to y=cos−1x: first find dxdy, then differentiate again to get dx2d2y.
Step 1 — the first derivative
The standard result for inverse cosine is
dxd(cos−1x)=−1−x21,−1<x<1.
It is the negative of the inverse-sine derivative, reflecting that cos−1x decreases as x increases.
Step 2 — differentiate again
Write the first derivative with a negative exponent so the chain rule is easy:
y′=−(1−x2)−1/2.
Differentiating,
y′′=−(−21)(1−x2)−3/2⋅(−2x),
where −21 comes from the power rule and −2x from the chain rule. Simplifying the signs and constants,
dx2d2(cos−1x)=−(1−x2)3/2x.
Note
This is the exact mirror of the inverse-sine result dx2d2(sin−1x)=(1−x2)3/2x — same shape, opposite sign — because their first derivatives already differ only by a sign.
Reading the result
At x=0: y′′=0, so the graph of cos−1x has an inflection at the origin.
For 0<x<1: y′′<0 (concave down); for −1<x<0: y′′>0 (concave up). …
Implicit differentiation gives (i); converting y=cos−1x back to x=cosy gives (ii) purely in terms of y. The OR part uses logarithmic differentiation and the parametric-derivative rule. …
Implicit differentiation gives (i); converting y=cos−1x back to x=cosy gives (ii) purely in terms of y. The OR part uses logarithmic differentiation and the parametric-derivative rule.
(i)sin2y+cos(xy)=k. Differentiate w.r.t. x:
2sinycosy⋅y′−sin(xy)⋅dxd(xy)=0
sin2y⋅y′−sin(xy)(y+xy′)=0
y′[sin2y−xsin(xy)]=ysin(xy)
y′=sin2y−xsin(xy)ysin(xy).
(ii)y=cos−1x⇒x=cosy, and 1−x2=1−cos2y=sin2y, so 1−x2=siny (as siny≥0 for y∈[0,π]). Then
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set 65/4/11 markMCQ
Q.If y=sin−1x, then (1−x2)dx2d2y is equal to : (A) xdxdy (B) −xdxdy (C) x2dxdy (D) −x2dxdy
›Reveal solutionSolution
We find the first and second derivatives of y=sin−1x. By simplifying the first derivative before taking the second, we arrive at a differential equation that directly gives the value of (1−x2)dx2d2y as xdxdy.
The problem asks us to find the value of the expression (1−x2)dx2d2y given that y=sin−1x. This requires us to calculate both the first derivative (dxdy) and the second derivative (dx2d2y) of y with respect to x. Once we have these, we will substitute them into the given expression and simplify.
A key strategy in problems involving higher-order derivatives of inverse trigonometric functions is to simplify the first derivative expression before differentiating it again. This often involves eliminating square roots or fractions, which makes the subsequent differentiation much cleaner and less prone to errors.
Find the first derivative, dxdy.
We are given the function y=sin−1x.
The standard derivative of sin−1x with respect to x is:
If y=sin−1x, then dxdy=1−x21.
So, our first derivative is:
dxdy=1−x21
Prepare for the second derivative by simplifying the first derivative expression.
To make the calculation of the second derivative easier, we can rearrange the expression for dxdy to remove the square root from the denominator. This is a common and effective technique.
Multiply both sides by 1−x2:
1−x2dxdy=1
Now, to eliminate the square root entirely, we square both sides of the equation:
(1−x2)2(dxdy)2=12
(1−x2)(dxdy)2=1
This form is much simpler to differentiate than the original fractional form.
Find the second derivative, dx2d2y.
We will now differentiate the equation (1−x2)(dxdy)2=1 with respect to x. We need to apply the product rule on the left side and the chain rule for (dxdy)2.
Let u=(1−x2) and v=(dxdy)2.
Then dxdu=−2x.
And dxdv=2(dxdy)dxd(dxdy)=2dxdydx2d2y.
Applying the product rule, dxd(uv)=udxdv+vdxdu: …