Q.Find the derivative of y=cos−1(3x) with respect to x.
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✓ Free question
Concept understanding — Derivatives of Inverse Trigonometric Functions
Inverse trigonometric functions (sin−1x,cos−1x,tan−1x,cot−1x,sec−1x,cosec−1x) are multi-valued in general, so a principal branch (a restricted domain and range) is fixed for each before differentiating. Each derivative is found by treating y= (inverse trig function of x) as x= (trig function of y), differentiating implicitly with respect to y, and using a Pythagorean identity to express the result back in terms of x. The six standard results are: dxdsin−1x=1−x21, dxdcos−1x=−1−x21, dxdtan−1x=1+x21, dxdcot−1x=−1+x21, dxdsec−1x=xx2−11 (for x>1; sign flips for x<−1), and dxdcosec−1x=−xx2−11 (for x>1; sign flips for x<−1). When the argument is a function f(x) rather than plain x, the chain rule multiplies each result by f′(x). Many composite inverse-trig expressions simplify dramatically using a trigonometric substitution (x=sinθ, x=tanθ, x=secθ, or x=acos2θ, guided by the shape of the expression) followed by a double- or triple-angle identity, collapsing the expression to a simple multiple of θ before differentiating.
Apply dxdcos−1(u)=−1−u21⋅dxdu with u=3x.
✓Final answer
dxdy=−1−9x23
With u=3x (so du/dx=3), Section 4's formula gives
dxdy=−1−(3x)21⋅3=−1−9x23,−31<x<31.
✓Final answer
dxdy=−1−9x23
Apply the standard cos−1 derivative formula with the chain-rule factor for the inner linear function u=3x; square u correctly as (3x)2=9x2.
Writing the denominator as 1−3x2 instead of 1−9x2 (forgetting to square the coefficient 3); dropping the negative sign from the cos−1 formula.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set V11 markMCQ
Q.The derivative of sin−1x exists in the interval
(a) [−1,1]
(b) (−1,1)
(c) R
(d) (2−π,2π)
›Reveal solutionSolution
The derivative 1−x21 exists only where 1−x2>0, i.e. on (−1,1); answer (b).
We have
dxdsin−1x=1−x21.
This is defined (finite) only when 1−x2>0, i.e. −1<x<1. At the endpoints x=±1 the denominator is 0, so the derivative does not exist there. Hence the interval is the open interval (−1,1).
✓Final answer
(b)(−1,1)
CBSE 2019Set ANNUAL1 markMCQ
Q.If y = tan^-1 [(5 - x) / (1 + 5x)], then value of dy/dx is
(a) -1/(1+x^2)
(b) 1/(1+x^2)
(c) 5
(d) 5/(1+x^2)
›Reveal solutionSolution
Recognize the expression as tan−1(5)−tan−1(x) via the tangent subtraction identity, then differentiate.
Recall tan(A−B)=1+tanAtanBtanA−tanB. With tanA=5 (constant) and tanB=x: