Question

Differentiate the given function w.r.t x, ${{x}^{x}}+{{x}^{a}}+{{a}^{x}}+{{a}^{a}}$ for some fixed a>0 and x>0.

Answer 1

Hint: In this question, we are given a function and we have to differentiate it. For this, we will use various properties of differentiation given as,
(1) The derivative of a sum of functions is equal to the sum of the derivative of functions.
(2) The derivative of a constant is equal to zero.
(3) Product rule of two functions $\dfrac{d}{dx}\left( u\cdot v \right)$ is given as $\dfrac{d}{dx}\left( u\cdot v \right)=\dfrac{udv}{dx}+\dfrac{vdu}{dx}$.
(4) Chain rule of $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)$ is given as $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)\cdot \dfrac{du}{dx}$.
(5) $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
(6) $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$.
Also we will use logarithm properties given as $\log {{a}^{b}}=b\log a$.

Complete step by step answer:
Here, we are given function as ${{x}^{x}}+{{x}^{a}}+{{a}^{x}}+{{a}^{a}}$ where a>0 and x>0.
Let the given function be equal to y. So, $y={{x}^{x}}+{{x}^{a}}+{{a}^{x}}+{{a}^{a}}$.
Let us suppose that ${{x}^{x}}=u,{{x}^{a}}=v,{{a}^{x}}=w\text{ and }{{a}^{a}}=s$.
Therefore, y = u+v+w+s differentiating y w.r.t x, we get:
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( u+v+w+s \right)\]
As we know, the derivative of the sum of a function is equal to the sum of the derivative of the function. Therefore,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}+\dfrac{ds}{dx}\cdots \cdots \cdots \left( 1 \right)\]
Now, we need to find values of $\dfrac{du}{dx},\dfrac{dv}{dx},\dfrac{dw}{dx},\dfrac{ds}{dx}$.
Now, $u={{x}^{x}}$ taking log on both sides, we get:
\[\Rightarrow \log u=\log {{x}^{x}}\]
As we know, $\log {{a}^{b}}=b\log a$ so we get:
\[\Rightarrow \log u=x\log x\]
Differentiating both sides w.r.t x, we get:
\[\Rightarrow \dfrac{d}{dx}\left( \log u \right)=\dfrac{d}{dx}\left( x\log x \right)\]
We know $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\text{ and }\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\cdot \dfrac{d}{dx}g\left( x \right)$ and product rule is given by $\dfrac{d}{dx}\left( u\cdot v \right)=\dfrac{udv}{dx}+\dfrac{vdu}{dx}$. Using these properties we get:
\[\begin{align}
  & \dfrac{1}{u}\cdot \dfrac{du}{dx}=x\dfrac{d}{dx}\left( \log x \right)+\log x\dfrac{d}{dx}\left( x \right) \\
 & \Rightarrow \dfrac{1}{u}\cdot \dfrac{du}{dx}=x\cdot \dfrac{1}{x}+\log x\cdot \left( 1 \right) \\
 & \Rightarrow \dfrac{1}{u}\cdot \dfrac{du}{dx}=1+\log x \\
\end{align}\]
Taking u on other side, we get:
\[\Rightarrow \dfrac{du}{dx}=u\left( 1+\log x \right)\]
Since $u={{x}^{x}}$ we get:
\[\Rightarrow \dfrac{du}{dx}={{x}^{x}}\left( 1+\log x \right)\cdots \cdots \cdots \left( 2 \right)\]
$v={{x}^{a}}$.
Differentiating both sides w.r.t x, we get:
\[\Rightarrow \dfrac{dv}{dx}=\dfrac{d}{dx}\left( {{x}^{a}} \right)\]
As we know, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ so we get:
\[\Rightarrow \dfrac{dv}{dx}=a{{x}^{a-1}}\cdots \cdots \cdots \left( 3 \right)\]
$w={{a}^{x}}$.
Taking log on both sides, we get:
\[\Rightarrow \log w=\log {{a}^{x}}\]
Since $\log {{a}^{b}}=b\log a$ we get:
\[\Rightarrow \log w=x\log a\]
Differentiating both sides w.r.t x, we get:
\[\Rightarrow \dfrac{1}{w}\cdot \dfrac{dw}{dx}=x\dfrac{d\log a}{dx}+\log a\dfrac{dx}{dx}\]
As we know, differentiation of constant term is zero, therefore we get:
\[\begin{align}
  & \Rightarrow \dfrac{1}{w}\cdot \dfrac{dw}{dx}=x\cdot 0+\log a\left( 1 \right) \\
 & \Rightarrow \dfrac{1}{w}\cdot \dfrac{dw}{dx}=\log a \\
\end{align}\]
Taking w on other side, we get:
\[\Rightarrow \dfrac{dw}{dx}=w\log a\]
Since $w={{a}^{x}}$ therefore,
\[\Rightarrow \dfrac{dw}{dx}={{a}^{x}}\log a\cdots \cdots \cdots \left( 4 \right)\]
$s={{a}^{a}}$.
Since 'a' is constant, therefore ${{a}^{a}}$ is also constant and derivatives of constant term is zero. Therefore, differentiating both sides w.r.t x, we get:
\[\begin{align}
  & \Rightarrow \dfrac{ds}{dx}=\dfrac{d}{dx}\left( {{a}^{a}} \right) \\
 & \Rightarrow \dfrac{ds}{dx}=0\cdots \cdots \cdots \left( 5 \right) \\
\end{align}\]
Putting values of (2), (3), (4), and (5) in equation (1) we get:
\[\begin{align}
  & \dfrac{dy}{dx}={{x}^{x}}\left( 1+\log x \right)+a{{x}^{a-1}}+{{a}^{x}}\log x+0 \\
 & \Rightarrow {{x}^{x}}\left( 1+\log x \right)+a{{x}^{a-1}}+{{a}^{x}}\log x \\
\end{align}\]

Hence, derivative of ${{x}^{x}}+{{x}^{a}}+{{a}^{x}}+{{a}^{a}}$ is ${{x}^{x}}\left( 1+\log x \right)+a{{x}^{a-1}}+{{a}^{x}}\log x$.

Note: Students should carefully find derivatives of all terms. They should keep in mind that, 'a' is constant here and so, its derivative is zero whereas u, v, w, s are functions, so we have to find $\dfrac{du}{dx},\dfrac{dv}{dx},\dfrac{dw}{dx},\dfrac{ds}{dx}$. While calculating differentiation of logarithmic functions, make sure that both $\dfrac{1}{u},\dfrac{du}{dx}$ terms should be written. Here, x>0 and a>0 are given because if they were negative, the function would move to the denominator and hence, the derivative would change.