Question

Differentiate $x^{\sin x}$ with respect to ${(\sin x)}^x$.

Answer 1

Hint:To solve this problem, we should know the basic concepts of differentiation with respect to a variable. In general, if we have to function g(x) and f(x). Then, to find the differentiation of g(x) with respect to f(x), we find the derivative of g(x) with respect to x and then find the derivative of f(x) with respect to x. Finally, we will divide these two results. In short, we have -
${\displaystyle \frac{d(g(x))}{d(f(x))}}={\displaystyle \frac{{\displaystyle \frac{d(g(x))}{dx}}}{{\displaystyle \frac{d(f(x))}{dx}}}}$
We will use the chain rule of differentiation and the product rule of differentiation along with this. We have, the product rule of differentiation as -
${\displaystyle \frac{d(p(x)q(x))}{dx}}=p(x){\displaystyle \frac{d(q(x))}{dx}}+q(x){\displaystyle \frac{d(p(x))}{dx}}$
The chain rule is given as –
Derivative of f(g(x)) is ${\displaystyle \frac{d(f(g(x)))}{dx}}\left({\displaystyle \frac{d(g(x))}{dx}}\right)$
We can now solve this problem by using these formulas.

Complete step-by-step answer:
Before solving this problem, we would first define the functions as g(x) = $x^{\sin x}$ and f(x) = ${(\sin x)}^x$. Now, since, we have to find ${\displaystyle \frac{d(g(x))}{d(f(x))}}={\displaystyle \frac{{\displaystyle \frac{d(g(x))}{dx}}}{{\displaystyle \frac{d(f(x))}{dx}}}}$, we first start by finding the derivative of g(x) with respect to x first. We have,
= ${\displaystyle \frac{d(g(x))}{dx}}$
= ${\displaystyle \frac{d(x^{\sin x})}{dx}}$
The above expression cannot be solved by normal properties of differentiation properties since the exponent of x is not a constant. Thus, to solve this type of problems, we have to take logarithm on both sides of g(x), thus, we have,
g(x) = $x^{\sin x}$
log(g(x)) = (sinx) log(x) -- (A)
Now, we use product rule of differentiation, this rule is given by –
${\displaystyle \frac{d(p(x)q(x))}{dx}}=p(x){\displaystyle \frac{d(q(x))}{dx}}+q(x){\displaystyle \frac{d(p(x))}{dx}}$
Now, we have,
Taking p(x) = sin x and q(x) = log(x), we have,
${\displaystyle \frac{d(\sin x(\log x))}{dx}}={\displaystyle \frac{\sin x}{x}}+(\log x)(\cos x)$ -- (1)
Now, to find the derivative of log(g(x)) with respect to x (in the LHS part of equation (A)), we need to use the chain rule of differentiation. According to this rule, derivative of any function f(g(x)) is ${\displaystyle \frac{d(f(g(x)))}{dx}}\left({\displaystyle \frac{d(g(x))}{dx}}\right)$. Thus, we have,
${\displaystyle \frac{d(\log (g(x))}{dx}}={\displaystyle \frac{1}{(g(x))}}{\displaystyle \frac{d(g(x))}{dx}}$ -- (2)

Substituting (1) and (2) in derivative of (A), we have,
${\displaystyle \frac{1}{(g(x))}}{\displaystyle \frac{d(g(x))}{dx}}={\displaystyle \frac{\sin x}{x}}+(\log x)(\cos x)$
${\displaystyle \frac{d(g(x))}{dx}}=(g(x))({\displaystyle \frac{\sin x}{x}}+(\log x)(\cos x))$ -- (B)
Now, we solve similarly for f(x), we have,
f(x) = ${(\sin x)}^x$
log(f(x)) = x log(sin(x)) -- (C)
Now, we use product rule of differentiation, this rule is given by –
${\displaystyle \frac{d(p(x)q(x))}{dx}}=p(x){\displaystyle \frac{d(q(x))}{dx}}+q(x){\displaystyle \frac{d(p(x))}{dx}}$
Now, first solving RHS of (C), we use product rule of differentiation and chain rule of differentiation, we have,
Taking p(x) = x and q(x) = log(sin(x)), we have,
${\displaystyle \frac{d(x(\log (\sin x))}{dx}}=x{\displaystyle \frac{\cos x}{\sin x}}+\log (\sin x)$ -- (3)
Now, solving the LHS part of (C), we use the chain rule. According to this rule, derivative of f(g(x)) is ${\displaystyle \frac{d(f(g(x)))}{dx}}\left({\displaystyle \frac{d(g(x))}{dx}}\right)$. Thus, using this on the LHS of equation (C), we have,
${\displaystyle \frac{d(\log (f(x))}{dx}}={\displaystyle \frac{1}{(f(x))}}{\displaystyle \frac{d(f(x))}{dx}}$ -- (4)

Substituting (3) and (4) as RHS and LHS respectively in derivative of (C), we have,
${\displaystyle \frac{1}{(f(x))}}{\displaystyle \frac{d(f(x))}{dx}}=x{\displaystyle \frac{\cos x}{\sin x}}+\log (\sin x)$
${\displaystyle \frac{d(f(x))}{dx}}=(f(x))(x{\displaystyle \frac{\cos x}{\sin x}}+\log (\sin x))$ -- (D)
Now, we divide (B) and (D), we have,
${\displaystyle \frac{d(g(x))}{d(f(x))}}={\displaystyle \frac{(g(x))({\displaystyle \frac{\sin x}{x}}+(\log x)(\cos x))}{(f(x))(x{\displaystyle \frac{\cos x}{\sin x}}+\log (\sin x))}}$
Substituting, the value of f(x) = ${(\sin x)}^x$ and g(x) = $x^{\sin x}$, we have,
${\displaystyle \frac{d(g(x))}{d(f(x))}}={\displaystyle \frac{(x^{\sin x})({\displaystyle \frac{\sin x}{x}}+(\log x)(\cos x))}{({(\sin x)}^x)(x{\displaystyle \frac{\cos x}{\sin x}}+\log (\sin x))}}$
The above expression is the final answer.

Note: When finding the derivative of any function, in most cases, we have to use the product rule of differentiation together with the chain rule of differentiation. This can be seen in the problem above where we had to use both the properties together such as for differentiating $x^{\sin x}$. Further, it is also useful to know about the basics of differentiation of basic functions like log(x), sin(x) and other functions since the differentiation of these functions are a must before we can use the chain rule or product rule of differentiation.