Question

Differentiate.\[{\sin ^2}\left( {\ln \,x} \right)\].

Answer 1

Hint: We will suppose the given value\[\left( {y = {{\sin }^2}\left( {\ln \,x} \right)} \right)\]. Differentiate the given value with respect to x.
\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]


Complete step by step solution:-
Let \[y = {\sin ^2}\left( {\ln x} \right)\]
Differentiate both side with respect to x, we will get
\[
  \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}{\sin ^2}\left( {\log x} \right) \\
  \dfrac{d}{{dx}}y = 2\sin \left( {\log x} \right)\dfrac{d}{{dx}}\sin \left( {\log x} \right) \\
  \dfrac{{dy}}{{dx}} = 2\sin \left( {\log x} \right)\cos \left( {\log x} \right)\dfrac{d}{{dx}}\left( {\log x} \right) \\
  \dfrac{{dy}}{{dx}} = 2\sin x\left( {\log x} \right)\cos x\left( {\log x} \right)\dfrac{1}{x}\dfrac{d}{{dx}}x \\
  \dfrac{{dy}}{{dx}} = 2\sin x\left( {\log x} \right)\cos x\left( {\log x} \right)\dfrac{1}{x} \times 1 \\
  \dfrac{{dy}}{{dx}} = 2\sin x\left( {\log x} \right)\cos x\left( {\log x} \right)\dfrac{1}{x} \\
  \dfrac{{dy}}{{dx}} = \dfrac{{2\sin x\left( {\log x} \right)\cos x\left( {\log x} \right)}}{x} \\
 \]
Additional information: Differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Some differentiation rule are:
(i) The constant rule: for any fixed real number $c$.\[\dfrac{d}{{dx}}\left\{ {c.f(x)} \right\} = c.\dfrac{d}{{dx}}\left\{ {f(x)} \right\}\]
(ii) The power rule: $\dfrac{d}{{dx}}\left\{ {{x^n}} \right\} = n{x^{n - 1}}$


Note: Students should follow product rule $\left[ {f\left( x \right)g\left( x \right)} \right]$ when we differentiate this value with respect to x then
$
  \dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] \\
   = g\left( n \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] \\
 $
DO not differentiate directly.