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Question: How do you differentiate \(f\left( x \right)={{e}^{x}}\ln x\)?...

How do you differentiate f(x)=exln⁡xf\left( x \right)={{e}^{x}}\ln x?

Explanation

Solution

We first define the multiplication rule and how the differentiation of function works. We take addition of these two different differentiated values. We take the dydx\dfrac{dy}{dx} altogether. We keep one function and differentiate the other one and then do the same thing with the other function. Then we take the addition to complete the formula.

Complete step-by-step solution:
We now discuss the multiplication process of two functions where f(x)=u(x)v(x)f\left( x \right)=u\left( x \right)v\left( x \right)
Differentiating f(x)=uvf\left( x \right)=uv, we get ddx[f(x)]=ddx[uv]=udvdx+vdudx\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}.
The above-mentioned rule is the multiplication rule. We apply that on f(x)=exln⁡xf\left( x \right)={{e}^{x}}\ln x. We assume the functions where u(x)=ex,v(x)=ln⁡xu\left( x \right)={{e}^{x}},v\left( x \right)=\ln x
We know that differentiation of u(x)=exu\left( x \right)={{e}^{x}} is u′(x)=ex{{u}^{'}}\left( x \right)={{e}^{x}} and differentiation of v(x)=ln⁡xv\left( x \right)=\ln x is v′(x)=1x{{v}^{'}}\left( x \right)=\dfrac{1}{x}. We now take differentiation on both parts of f(x)=exln⁡xf\left( x \right)={{e}^{x}}\ln x and get ddx[f(x)]=ddx[exln⁡x]\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{e}^{x}}\ln x \right].
We place the values of u′(x)=ex{{u}^{'}}\left( x \right)={{e}^{x}} and v′(x)=1x{{v}^{'}}\left( x \right)=\dfrac{1}{x} to get
ddx[exln⁡x]=exddx(ln⁡x)+(ln⁡x)ddx(ex)\dfrac{d}{dx}\left[ {{e}^{x}}\ln x \right]={{e}^{x}}\dfrac{d}{dx}\left( \ln x \right)+\left( \ln x \right)\dfrac{d}{dx}\left( {{e}^{x}} \right).
We take all the dydx\dfrac{dy}{dx} forms altogether to get

& \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{e}^{x}}\ln x \right] \\\ & \Rightarrow {{f}^{'}}\left( x \right)={{e}^{x}}\times \dfrac{1}{x}+\left( \ln x \right)\left( {{e}^{x}} \right)=\dfrac{{{e}^{x}}}{x}+{{e}^{x}}\ln x \\\ & \Rightarrow {{f}^{'}}\left( x \right)={{e}^{x}}\left( \ln x+\dfrac{1}{x} \right) \\\ \end{aligned}$$ **Therefore, differentiation of $f\left( x \right)={{e}^{x}}\ln x$ is $${{e}^{x}}\left( \ln x+\dfrac{1}{x} \right)$$.** **Note:** We need to remember that in the chain rule $$\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}$$, we aren’t cancelling out the part $$d\left[ h\left( x \right) \right]$$. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate. The rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts.