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Question: If \[y = \sec ({\tan ^{ - 1}}x)\] then \[\dfrac{{dy}}{{dx}}\] at \[x = 1\] is equal to A. \[\dfrac...

If y=sec(tan1x)y = \sec ({\tan ^{ - 1}}x) then dydx\dfrac{{dy}}{{dx}} at x=1x = 1 is equal to
A. 12\dfrac{1}{{\sqrt 2 }}
B. 12\dfrac{1}{2}
C. 11
D. 2\sqrt 2

Explanation

Solution

Hint : In mathematics , the inverse trigonometric functions are the inverse functions of the trigonometric functions. Specifically, inverse trigonometric functions are the inverses of the sine, the cosine, the tangent, the cotangent, the secant, and the cosecant functions and are used to obtain an angle from any of the angle's trigonometric ratios.

Complete step-by-step solution :
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). 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. 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.
We are given y=sec(tan1x)y = \sec ({\tan ^{ - 1}}x)
Now on differentiating both the sides with respect to variable xx we have
dydx=sec(tan1x)tan(tan1x)11+x2\dfrac{{dy}}{{dx}} = \sec ({\tan ^{ - 1}}x)\tan ({\tan ^{ - 1}}x)\dfrac{1}{{1 + {x^2}}}
We know that the trigonometric function and its inverse trigonometric part gets cancelled or compensated. Therefore we get
dydx=sec(tan1x)(x)11+x2\dfrac{{dy}}{{dx}} = \sec ({\tan ^{ - 1}}x)(x)\dfrac{1}{{1 + {x^2}}}
Putting the value x=1x = 1 we get
dydx=sec(tan11)11+1\dfrac{{dy}}{{dx}} = \sec ({\tan ^{ - 1}}1)\dfrac{1}{{1 + 1}}
Which on simplification becomes
dydx=sec(π4)12\dfrac{{dy}}{{dx}} = \sec \left( {\dfrac{\pi }{4}} \right)\dfrac{1}{2}
Which on further simplification becomes
dydx=(12)(2)=12\dfrac{{dy}}{{dx}} = \left( {\dfrac{1}{2}} \right)\left( {\sqrt 2 } \right) = \dfrac{1}{{\sqrt 2 }}
Therefore, the correct answer is Option (A).

Note: Inverse trigonometric functions are the inverses of the sine, the cosine, the tangent, the cotangent, the secant, and the cosecant functions and are used to obtain an angle from any of the angle's trigonometric ratios.While taking derivative of a function , always remember to use chain rule.