Differentiate \(tan^{-1}(\frac{\sqrt{1+x^2}-1}{x}) \,w.r.t\,\,cos^{-1}(\frac{\sqrt(1+\sqrt{1+x^2})}{2\sqrt({i}+x^2)})\)
List-I | List-II |
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The derivative of \( \log_e x \) with respect to \( \frac{1}{x} \) at \( x = 5 \) is | (I) -5 |
If \( x^3 + x^2y + xy^2 - 21x = 0 \), then \( \frac{dy}{dx} \) at \( (1, 1) \) is | (II) -6 |
If \( f(x) = x^3 \log_e \frac{1}{x} \), then \( f'(1) + f''(1) \) is | (III) 5 |
If \( y = f(x^2) \) and \( f'(x) = e^{\sqrt{x}} \), then \( \frac{dy}{dx} \) at \( x = 0 \) is | (IV) 0 |
List-I (Function) | List-II (Derivative w.r.t. x) | |
---|---|---|
(A) \( \frac{5^x}{\ln 5} \) | (I) \(5^x (\ln 5)^2\) | |
(B) \(\ln 5\) | (II) \(5^x \ln 5\) | |
(C) \(5^x \ln 5\) | (III) \(5^x\) | |
(D) \(5^x\) | (IV) 0 |
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)