Question

Differentiate the function ${{e}^{x}}+{{e}^{{{x}^{2}}}}+....+{{e}^{{{x}^{5}}}}$ w.r.t x.

Answer 1

Hint: We start solving the problem by equating the given function to y. We then make use of the property $\dfrac{d}{dx}\left( a\left( x \right)+b\left( x \right)+......+g\left( x \right) \right)=\dfrac{d}{dx}\left( a\left( x \right) \right)+\dfrac{d}{dx}\left( b\left( x \right) \right)+......+\dfrac{d}{dx}\left( g\left( x \right) \right)$ to proceed through the problem. We then make use of the properties $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ and $\dfrac{d}{dx}\left( {{e}^{f\left( x \right)}} \right)={{e}^{f\left( x \right)}}\dfrac{d\left( f\left( x \right) \right)}{dx}$ to proceed further into the problem. We then make use of the property $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and make the necessary arrangements to get the required answer.

Complete step-by-step solution
According to the problem, we are asked to differentiate the function ${{e}^{x}}+{{e}^{{{x}^{2}}}}+....+{{e}^{{{x}^{5}}}}$ w.r.t x.
Let us assume $y={{e}^{x}}+{{e}^{{{x}^{2}}}}+....+{{e}^{{{x}^{5}}}}$ and differentiate both sides w.r.t x.
So, we get $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{e}^{x}}+{{e}^{{{x}^{2}}}}+....+{{e}^{{{x}^{5}}}} \right)$ ---(1).
We know that $\dfrac{d}{dx}\left( a\left( x \right)+b\left( x \right)+......+g\left( x \right) \right)=\dfrac{d}{dx}\left( a\left( x \right) \right)+\dfrac{d}{dx}\left( b\left( x \right) \right)+......+\dfrac{d}{dx}\left( g\left( x \right) \right)$, let us use this result in equation (1).
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{x}} \right)}{dx}+\dfrac{d\left( {{e}^{{{x}^{2}}}} \right)}{dx}+\dfrac{d\left( {{e}^{{{x}^{3}}}} \right)}{dx}+\dfrac{d\left( {{e}^{{{x}^{4}}}} \right)}{dx}+\dfrac{d\left( {{e}^{{{x}^{5}}}} \right)}{dx}$ ---(2).
We know that $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ and $\dfrac{d}{dx}\left( {{e}^{f\left( x \right)}} \right)={{e}^{f\left( x \right)}}\dfrac{d\left( f\left( x \right) \right)}{dx}$, let us use these results in equation (2)
$\Rightarrow \dfrac{dy}{dx}={{e}^{x}}+{{e}^{{{x}^{2}}}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+{{e}^{{{x}^{3}}}}\dfrac{d\left( {{x}^{3}} \right)}{dx}+{{e}^{{{x}^{4}}}}\dfrac{d\left( {{x}^{4}} \right)}{dx}+{{e}^{{{x}^{5}}}}\dfrac{d\left( {{x}^{5}} \right)}{dx}$ ---(3).
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, we use this result in equation (3).
$\Rightarrow \dfrac{dy}{dx}={{e}^{x}}+{{e}^{{{x}^{2}}}}\left( 2x \right)+{{e}^{{{x}^{3}}}}\left( 3{{x}^{2}} \right)+{{e}^{{{x}^{4}}}}\left( 4{{x}^{3}} \right)+{{e}^{{{x}^{5}}}}\left( 5{{x}^{4}} \right)$.
$\Rightarrow \dfrac{dy}{dx}={{e}^{x}}+2x{{e}^{{{x}^{2}}}}+3{{x}^{2}}{{e}^{{{x}^{3}}}}+4{{x}^{3}}{{e}^{{{x}^{4}}}}+5{{x}^{4}}{{e}^{{{x}^{5}}}}$.
So, we have found the result of differentiation of the function ${{e}^{x}}+{{e}^{{{x}^{2}}}}+....+{{e}^{{{x}^{5}}}}$ w.r.t x as ${{e}^{x}}+2x{{e}^{{{x}^{2}}}}+3{{x}^{2}}{{e}^{{{x}^{3}}}}+4{{x}^{3}}{{e}^{{{x}^{4}}}}+5{{x}^{4}}{{e}^{{{x}^{5}}}}$.
$\therefore$ $\dfrac{d}{dx}\left( {{e}^{x}}+{{e}^{{{x}^{2}}}}+....+{{e}^{{{x}^{5}}}} \right)={{e}^{x}}+2x{{e}^{{{x}^{2}}}}+3{{x}^{2}}{{e}^{{{x}^{3}}}}+4{{x}^{3}}{{e}^{{{x}^{4}}}}+5{{x}^{4}}{{e}^{{{x}^{5}}}}$.

Note: We need to perform calculations carefully in each step to avoid confusion and mistakes. We should not be confused especially while performing chain rule for the exponential function ${{e}^{{{x}^{n}}}}$. We can also solve this problem by making use of the fact that the derivative of a function $f\left( x \right)$ is defined as ${{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$, which may involve a tricky calculation. We know that the exponential function is continuous and differentiable so we need not check it again. Similarly, we can expect problems to find the derivative of the function $\ln x+\ln \left( \ln x \right)+......+\ln \left( \ln \left( \ln \left( \ln \left( \ln \left( \ln x \right) \right) \right) \right) \right)$ w.r.t x.