Find the derivative dy/dx of y = sin^-1(2x/(1+x^2)) at x = 1.

Find the derivative $\frac{d y}{d x}$ of $y = sin^{- 1} (\frac{2 x}{1 + x ^{2}})$ at $x = 1$ .

medium 2 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

Derivative does not exist at

x = 1

The correct answer is Derivative does not exist at $x = 1$ — Therefore $\frac{d y}{d x}$ is undefined (does not exist) at $x = 1$ .

Let $u (x) = \frac{2 x}{1 + x ^{2}}$ . Then $y = sin^{- 1} u$ .
Compute $\frac{d u}{d x} = \frac{2 ( 1 + x ^{2} ) - 2 x ( 2 x )}{( 1 + x ^{2} ) ^{2}} = \frac{2 - 2 x ^{2}}{( 1 + x ^{2} ) ^{2}} = \frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{2}}$ .
Derivative of $sin^{- 1} u$ is $\frac{d u / d x}{1 - u ^{2}}$ .
Evaluate $u$ at $x = 1$ : $u (1) = \frac{2 \cdot 1}{1 + 1} = 1$ .
Then $1 - u^{2} = 1 - 1^{2} = 0$ . The denominator of the derivative formula becomes zero.
Since the denominator approaches zero while the numerator $\frac{d u}{d x}$ is finite ( $\frac{d u}{d x}_{x = 1} = 0$ ), the limit is not finite; the function has a vertical tangent.
Therefore $\frac{d y}{d x}$ is undefined (does not exist) at $x = 1$ .

Common mistake: Substituting

x = 1

directly into the formula without checking that

1 - u^{2}

becomes zero.

Concept tested: differentiation of inverse sine composite function