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Question: Differentiate \({\operatorname{sech} ^{ - 1}}x\) with respect to x, by first writing \(x = \operator...

Differentiate sech⁡−1x{\operatorname{sech} ^{ - 1}}x with respect to x, by first writing x=sech⁡yx = \operatorname{sech} y.

Explanation

Solution

Hint: To solve this question, we will use the result obtained by differentiation of hyperbolic functions (cosh⁡x\cosh x, sinh⁡x\sinh x,etc). Also, we will use some properties of hyperbolic functions.

Complete step-by-step answer:
Now, we are given the function y=sech⁡−1xy = {\operatorname{sech} ^{ - 1}}x. Rewriting this function, we get
x=sech⁡yx = \operatorname{sech} y … (1)
Now, from hyperbolic functions, we know that sech⁡x=1cosh⁡y\operatorname{sech} x = \dfrac{1}{{\cosh y}}. So, equation (1) becomes,
x=1cosh⁡yx = \dfrac{1}{{\cosh y}}
xcosh⁡y=1x\cosh y = 1
Now, differentiating the above function on both sides, with respect to x.
d(xcosh⁡y)dx=d(1)dx\dfrac{{d(x\cosh y)}}{{dx}} = \dfrac{{d(1)}}{{dx}} … (2)
Now, as 1 is constant, so its differentiation is equal to zero, i.e. d(1)dx=0\dfrac{{d(1)}}{{dx}} = 0.
Also, to differentiate the left-hand side term, we will use the product-rule of differentiation.
Product rule of differentiation for a function y=vxy = vx is dydx=vd(x)dx+xdvdx\dfrac{{dy}}{{dx}} = v\dfrac{{d(x)}}{{dx}} + x\dfrac{{dv}}{{dx}}
So, using this rule, we get d(xcosh⁡y)dx=1(cosh⁡y)+xd(cosh⁡y)dx\dfrac{{d(x\cosh y)}}{{dx}} = 1(\cosh y) + x\dfrac{{d(\cosh y)}}{{dx}}
Now, d(cosh⁡x)dx=sinh⁡x\dfrac{{d(\cosh x)}}{{dx}} = \sinh x
Therefore, d(xcosh⁡y)dx=xsinh⁡ydydx+cosh⁡y\dfrac{{d(x\cosh y)}}{{dx}} = x\sinh y\dfrac{{dy}}{{dx}} + \cosh y
So, equation (2) becomes,
xsinh⁡ydydx+cosh⁡y=0x\sinh y\dfrac{{dy}}{{dx}} + \cosh y = 0
dydx=−cosh⁡yxsinh⁡y\dfrac{{dy}}{{dx}} = - \dfrac{{\cosh y}}{{x\sinh y}}
As, we know cosh⁡ysinh⁡y=1tanh⁡y\dfrac{{\cosh y}}{{\sinh y}} = \dfrac{1}{{\tanh y}}
Therefore, dydx=−1xtanh⁡y\dfrac{{dy}}{{dx}} = - \dfrac{1}{{x\tanh y}}
Also, from hyperbolic functions, we have tanh⁡2x+sech⁡2x=1{\tanh ^2}x + {\operatorname{sech} ^2}x = 1
Therefore, we can write tanh⁡x=1−sech⁡2x\tanh x = \sqrt {1 - \operatorname{sech} {}^2x}
Putting this value in dydx=−1xtanh⁡y\dfrac{{dy}}{{dx}} = - \dfrac{1}{{x\tanh y}}, we get
dydx=−1x1−sech⁡2y\dfrac{{dy}}{{dx}} = - \dfrac{1}{{x\sqrt {1 - {{\operatorname{sech} }^2}y} }}
But, x=sech⁡yx = \operatorname{sech} y
Therefore, dydx=−1x1−x2\dfrac{{dy}}{{dx}} = - \dfrac{1}{{x\sqrt {1 - {x^2}} }}

Note: Whenever we come up with such types of questions, we will use some properties of hyperbolic functions. Also, various results of differentiation of hyperbolic functions are useful in solving such types of problems. Always, write the final answer in terms of the relation given in the question, like in this question we are given the relation x=sech⁡yx = \operatorname{sech} yand we write the final answer in terms of x.