Question

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

Answer 1

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.