Question

Differentiate \[{x^{\sin x}}\],\[x > 0\] with respect to x.

Answer 1

Hint: We use the concept of log in this question. Write the value given equal to a different variable and apply logarithm on both sides of the equation. Use the property of log and break the right hand side. Differentiate using product rule of differentiation.
* \[\log {m^n} = n\log m\]
* \[\log m + \log n = \log mn\]
* Product rule of differentiation:\[\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}(b) + b\dfrac{d}{{dx}}(a)\]
* Chain rule of differentiation:\[\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)\]

Complete step-by-step answer:
We have to find\[\dfrac{d}{{dx}}({x^{\sin x}})\]
Here the function is \[{x^{\sin x}}\]
Let us assume the function equal to a variable y.
Let\[y = {x^{\sin x}}\] … (1)
Now we apply log function on both sides of the equation.
\[ \Rightarrow \log y = \log ({x^{\sin x}})\]
Use the property\[\log {m^n} = n\log m\]where m is x and n is x.
\[ \Rightarrow \log y = \sin x(\log x)\]
Now differentiate both sides of the equation with respect to x
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {\sin x(\log x)} \right)\]
Apply chain rule of differentiation in LHs of the equation
Chain rule gives us\[\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)\].
Here\[g(f(x)) = \log (y),f(x) = y\], then the equation becomes
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) \times \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin x(\log x)} \right)\]
We know \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]
\[ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin x(\log x)} \right)\]
Now apply product rule of differentiation in RHS of the equation
Product rule gives us\[\dfrac{d}{{dx}}\left( {ab} \right) = a\dfrac{d}{{dx}}(b) + b\dfrac{d}{{dx}}(a)\]
Here\[a = \sin x,b = \log x\], then the equation becomes
\[ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \sin x\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}\sin x\]
Substitute the values\[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]and\[\dfrac{d}{{dx}}\sin = \cos x\]in RHS of the equation
\[ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \sin x \times \dfrac{1}{x} + \log x \times \cos x\]
Multiply the terms in RHS of the equation
\[ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \dfrac{{\sin x}}{x} + \log x.\cos x\]
Multiply both sides of the equation by y
\[ \Rightarrow y \times \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = y \times \left( {\dfrac{{\sin x}}{x} + \log x.\cos x} \right)\]
Cancel same function from numerator and denominator in LHS of the equation
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\left( {\dfrac{{\sin x}}{x} + \log x.\cos x} \right)\]
Substitute the value of \[y = {x^{\sin x}}\]from equation (1)
\[ \Rightarrow \dfrac{d}{{dx}}({x^{\sin x}}) = {x^{\sin x}}\left( {\dfrac{{\sin x}}{x} + \log x.\cos x} \right)\]
Take LCM in right side of the equation
\[ \Rightarrow \dfrac{d}{{dx}}({x^{\sin x}}) = {x^{\sin x}}\left( {\dfrac{{\sin x + x\log x\cos x}}{x}} \right)\]
Since we can write \[\dfrac{1}{x} = {x^{ - 1}}\]
\[ \Rightarrow \dfrac{d}{{dx}}({x^{\sin x}}) = {x^{\sin x}}.{x^{ - 1}}\left( {\sin x + x\log x\cos x} \right)\]
We have x as same base so we can add the powers
\[ \Rightarrow \dfrac{d}{{dx}}({x^{\sin x}}) = {x^{\sin x - 1}}\left( {\sin x + x\log x\cos x} \right)\]

\[\therefore \]Differentiation of \[{x^{\sin x}}\]with respect to x is \[{x^{\sin x - 1}}\left( {\sin x + x\log x\cos x} \right)\]

Note:
 Students many times make the mistake of writing the final answer of differentiation without shifting or removing the value of y from the left hand side of the equation. Keep in mind we only need the value of differentiation of y with respect to x i.e. differentiation of the given function with respect to x.