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Question: If \[y = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}}x + \cos e{c^{ - 1}}x\] then \[\dfrac{{dy...

If y=tan⁡−1x+cot⁡−1x+sec⁡−1x+cos⁡ec−1xy = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}}x + \cos e{c^{ - 1}}x then dydx\dfrac{{dy}}{{dx}} is equal to
A. (x2−1)(x2+1)\dfrac{{\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)}}
B. π\pi
C. 00
D. 11

Explanation

Solution

The process of finding out the derivative of a function with respect to a given variable is called differentiation. The reverse process is called anti differentiation. The fundamental theorem of calculus relates anti differentiation with integration. Here in this question we will require derivatives of inverse trigonometric functions .

Complete step by step answer:
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value. For this reason, the derivative is often described as the "instantaneous rate of change", the ratio of the instantaneous change in the dependent variable to that of the independent variable.
FORMULAS USED IN THE SOLUTION TO THIS QUESTION :
Derivative of tan⁡−1x{\tan ^{ - 1}}x is dydxtan⁡−1x=11+x2\dfrac{{dy}}{{dx}}{\tan ^{ - 1}}x = \dfrac{1}{{1 + {x^2}}}
Derivative of cot⁡−1x{\cot ^{ - 1}}x is dydxcot⁡−1x=−11+x2\dfrac{{dy}}{{dx}}{\cot ^{ - 1}}x = - \dfrac{1}{{1 + {x^2}}}
Derivative of sec⁡−1x{\sec ^{ - 1}}x is dydxsec⁡−1x=1∣x∣x2−1\dfrac{{dy}}{{dx}}{\sec ^{ - 1}}x = \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}
Derivative of cos⁡ec−1x\cos e{c^{ - 1}}x is dydxcos⁡ec−1x=−1∣x∣x2−1\dfrac{{dy}}{{dx}}\cos e{c^{ - 1}}x = - \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}
Now consider the given equation ,
y=tan⁡−1x+cot⁡−1x+sec⁡−1x+cos⁡ec−1xy = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}}x + \cos e{c^{ - 1}}x
On differentiating both the sides with respect to the variable xx we get ,
dydx=11+x2−11+x2+1∣x∣x2−1−1∣x∣x2−1\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {x^2}}} - \dfrac{1}{{1 + {x^2}}} + \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }} - \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}
Therefore we get ,
dydx=0\dfrac{{dy}}{{dx}} = 0

So, the correct answer is “Option C”.

Note: Keep in mind all the formulas of differentiation. Note that integration is known as anti differentiation . You should be well versed with all the formulas of differentiation. Here in this question we will require derivatives of inverse trigonometric functions .