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Differentiate the given function w.r.t x:
${\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}},\dfrac{\pi }{4} < x < \dfrac{{3\pi }}{4}$

Last updated date: 30th May 2024
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Hint: As the function is in the form of variable to the power of variable we apply log on both sides of the equation and then differentiate.

Complete step-by-step answer:
Let $y = {\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}}$
Take log both the side
$ \Rightarrow \log y = \log {\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}}$
We know that $\log {a^b} = b\log a$
$ \Rightarrow \log y = \left( {\sin x - \cos x} \right)\log \left( {\sin x - \cos x} \right)$
Now differentiate both the side w.r.t x
Here we use chain rule of differentiation
Differentiation of sinx wrt x is cosx
Differentiation of cosx wrt x is -sinx
Differentiation of logx wrt x is $\dfrac{1}{x}$

$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\sin x - \cos x} \right)\dfrac{{d\log \left( {\sin x - \cos x} \right)}}{{dx}} + \log \left( {\sin x - \cos x} \right)\dfrac{{d\left( {sinx - \cos x} \right)}}{{dx}}$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\sin x - \cos x} \right)\dfrac{1}{{\left( {\sin x - \cos x} \right)}}\dfrac{{d\left( {\sin x - \cos x} \right)}}{{dx}} + \log \left( {\sin x - \cos x} \right)\left( {\cos x + \sin x} \right)$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\sin x - \cos x} \right)\dfrac{1}{{\left( {\sin x - \cos x} \right)}}\left( {\cos x + \sin x} \right) + \log \left( {\sin x - \cos x} \right)\left( {\cos x + \sin x} \right)$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {\cos x + \sin x} \right) + \log \left( {\sin x - \cos x} \right)\left( {\cos x + \sin x} \right)$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {1 + log\left( {\sin x - \cos x} \right)} \right)\left( {\cos x + \sin x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = y\left( {1 + log\left( {\sin x - \cos x} \right)} \right)\left( {\cos x + \sin x} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\sin x - \cos x} \right)^{\left( {\sin x - \cos x} \right)}}\left( {1 + log\left( {\sin x - \cos x} \right)} \right)\left( {\cos x + \sin x} \right)$
So this is your required answer.

Note: In this type of question always take log on both sides then solve, If there is any function in log then after applying differentiation of log function is again differentiated w.r.t the variable.