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Question: How do you differentiate \[y={{x}^{\sqrt{x}}}\]?...

How do you differentiate y=xxy={{x}^{\sqrt{x}}}?

Explanation

Solution

This question is from the topic of calculus. In solving this question, we will first put log base e (that is ln⁡\ln ) to the both sides of the equation. After that, we will differentiate both sides of the equation. We will use the formula d(ln⁡x)=(1x)dxd\left( \ln x \right)=\left( \dfrac{1}{x} \right)dx and d(xn)=(nxn−1)dxd\left( {{x}^{n}} \right)=\left( n{{x}^{n-1}} \right)dx for the further differentiation. After that, we will solve the further question and find the value of dydx\dfrac{dy}{dx} that will be our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to differentiate y=xxy={{x}^{\sqrt{x}}}. Or, we can say that we have to find the value of dydx\dfrac{dy}{dx} using the equation y=xxy={{x}^{\sqrt{x}}}.
The term which we have to solve is
y=xxy={{x}^{\sqrt{x}}}
Now, putting ln⁡\ln (that is log base ‘e’ or we can say log⁡e{{\log }_{e}}) to the both side of the equation, we get
⇒ln⁡y=ln⁡xx\Rightarrow \ln y=\ln {{x}^{\sqrt{x}}}
Now, using the formula of logarithms that is ln⁡xa=aln⁡x\ln {{x}^{a}}=a\ln x, we can write the above equation as
⇒ln⁡y=xln⁡x\Rightarrow \ln y=\sqrt{x}\ln x
Now, differentiating both side of equation, we can write
⇒d(ln⁡y)=d(xln⁡x)\Rightarrow d\left( \ln y \right)=d\left( \sqrt{x}\ln x \right)
Now, using the formula of product rule that is d(u.v)=vd(u)−ud(v)d\left( u.v \right)=vd\left( u \right)-ud\left( v \right), we can write the above equation as
⇒d(ln⁡y)=ln⁡x⋅d(x)+x⋅d(ln⁡x)\Rightarrow d\left( \ln y \right)=\ln x\centerdot d\left( \sqrt{x} \right)+\sqrt{x}\centerdot d\left( \ln x \right)
⇒d(ln⁡y)=ln⁡x⋅d(x12)+x⋅d(ln⁡x)\Rightarrow d\left( \ln y \right)=\ln x\centerdot d\left( {{x}^{\dfrac{1}{2}}} \right)+\sqrt{x}\centerdot d\left( \ln x \right)
Now, using the formula of differentiation that is d(xn)=(nxn−1)dxd\left( {{x}^{n}} \right)=\left( n{{x}^{n-1}} \right)dx, we can write
⇒d(ln⁡y)=ln⁡x⋅(12x12−1)dx+x⋅d(ln⁡x)\Rightarrow d\left( \ln y \right)=\ln x\centerdot \left( \dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \right)dx+\sqrt{x}\centerdot d\left( \ln x \right)
⇒d(ln⁡y)=ln⁡x⋅(12x−12)dx+x⋅d(ln⁡x)\Rightarrow d\left( \ln y \right)=\ln x\centerdot \left( \dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \right)dx+\sqrt{x}\centerdot d\left( \ln x \right)
Now, using the formula of differentiation that is d(ln⁡x)=(1x)dxd\left( \ln x \right)=\left( \dfrac{1}{x} \right)dx, we can write the above equation as
⇒(1y)dy=ln⁡x⋅(12x−12)dx+x⋅(1x)dx\Rightarrow \left( \dfrac{1}{y} \right)dy=\ln x\centerdot \left( \dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \right)dx+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)dx
The above equation can also be written as
⇒(1y)dy=ln⁡x⋅(12(x12))dx+x⋅(1x)dx\Rightarrow \left( \dfrac{1}{y} \right)dy=\ln x\centerdot \left( \dfrac{1}{2\left( {{x}^{\dfrac{1}{2}}} \right)} \right)dx+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)dx
⇒(1y)dy=ln⁡x⋅(12x)dx+x⋅(1x)dx\Rightarrow \left( \dfrac{1}{y} \right)dy=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)dx+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)dx
Now dividing ‘dx’ to both side of the equation, we can write
⇒(1y)dydx=ln⁡x⋅(12x)+x⋅(1x)\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)
The above equation can also be written as
⇒(1y)dydx=ln⁡x⋅(12x)+x⋅(1x⋅x)\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)+\sqrt{x}\centerdot \left( \dfrac{1}{\sqrt{x}\centerdot \sqrt{x}} \right)
⇒(1y)dydx=ln⁡x⋅(12x)+(1x)\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)+\left( \dfrac{1}{\sqrt{x}} \right)
The above equation can also be written as
⇒(1y)dydx=ln⁡x2(1x)+(1x)\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\dfrac{\ln x}{2}\left( \dfrac{1}{\sqrt{x}} \right)+\left( \dfrac{1}{\sqrt{x}} \right)
Taking 1x\dfrac{1}{\sqrt{x}} as common in the right side of the equation, we can write
⇒(1y)dydx=(1x)(ln⁡x2+1)\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\left( \dfrac{1}{\sqrt{x}} \right)\left( \dfrac{\ln x}{2}+1 \right)
⇒dydx=y(1x)(ln⁡x2+1)\Rightarrow \dfrac{dy}{dx}=y\left( \dfrac{1}{\sqrt{x}} \right)\left( \dfrac{\ln x}{2}+1 \right)
Now, putting the value of y in the above equation, we can write
⇒dydx=xx(1x)(ln⁡x2+1)\Rightarrow \dfrac{dy}{dx}={{x}^{\sqrt{x}}}\left( \dfrac{1}{\sqrt{x}} \right)\left( \dfrac{\ln x}{2}+1 \right)
⇒dydx=xx(x−12)(ln⁡x2+1)\Rightarrow \dfrac{dy}{dx}={{x}^{\sqrt{x}}}\left( {{x}^{-\dfrac{1}{2}}} \right)\left( \dfrac{\ln x}{2}+1 \right)
Now, using the formula xaxb=xa+b{{x}^{a}}{{x}^{b}}={{x}^{a+b}}, we can write
⇒dydx=xx−12(ln⁡x2+1)\Rightarrow \dfrac{dy}{dx}={{x}^{\sqrt{x}-\dfrac{1}{2}}}\left( \dfrac{\ln x}{2}+1 \right)
Or, we can write the above equation as
⇒dydx=x−12+x(ln⁡x+22)\Rightarrow \dfrac{dy}{dx}={{x}^{-\dfrac{1}{2}+\sqrt{x}}}\left( \dfrac{\ln x+2}{2} \right)
⇒dydx=12x−12+x(ln⁡x+2)\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}+\sqrt{x}}}\left( \ln x+2 \right)
Hence, we have done differentiation now. The differentiation is dydx=12x−12+x(ln⁡x+2)\dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}+\sqrt{x}}}\left( \ln x+2 \right)

Note:
We should have a better knowledge in the topic of pre-calculus for solving this type of question easily. We should the following formulas:
Product rule of differentiation: d(u.v)=vd(u)−ud(v)d\left( u.v \right)=vd\left( u \right)-ud\left( v \right)
Logarithm formula: ln⁡xa=aln⁡x\ln {{x}^{a}}=a\ln x
d(ln⁡x)=(1x)dxd\left( \ln x \right)=\left( \dfrac{1}{x} \right)dx
d(xn)=(nxn−1)dxd\left( {{x}^{n}} \right)=\left( n{{x}^{n-1}} \right)dx
x=x12\sqrt{x}={{x}^{\dfrac{1}{2}}}
1x=x−(12)\dfrac{1}{\sqrt{x}}={{x}^{-\left( \dfrac{1}{2} \right)}}
xaxb=xa+b{{x}^{a}}{{x}^{b}}={{x}^{a+b}}