Question

What is the derivative of $y=\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}$?

Answer 1

To solve the given question, we should know how to differentiate composite functions. The composite functions are functions of the form $f\left( g(x) \right)$, their derivative is found as, $\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}$. We should also know the quotient rule of differentiation which states that $\dfrac{d\left( \dfrac{f(x)}{g(x)} \right)}{dx}=\dfrac{\dfrac{d\left( f(x) \right)}{dx}g(x)-f(x)\dfrac{d\left( g(x) \right)}{dx}}{{{\left( g(x) \right)}^{2}}}$. The derivatives of functions ${{e}^{x}},\dfrac{1}{x}\And {{x}^{2}}$ are ${{e}^{x}},-\dfrac{1}{{{x}^{2}}}\And 2x$.

Complete step by step solution:
We know that the derivative of the composite function is evaluated as $\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{d\left( f\left( g(x) \right) \right)}{d\left( g(x) \right)}\dfrac{d\left( g(x) \right)}{dx}$. First, we need to evaluate the derivative of function ${{e}^{\dfrac{1}{x}}}$ using this as, $\dfrac{d\left( {{e}^{\dfrac{1}{x}}} \right)}{dx}=\dfrac{d\left( {{e}^{\dfrac{1}{x}}} \right)}{d\left( \dfrac{1}{x} \right)}\dfrac{d\left( \dfrac{1}{x} \right)}{dx}$.
We know the derivative of ${{e}^{x}}$ is ${{e}^{x}}$, and derivative of $\dfrac{1}{x}$ is $-\dfrac{1}{{{x}^{2}}}$. Using this we can evaluate the above derivative as,
$\dfrac{d\left( {{e}^{\dfrac{1}{x}}} \right)}{dx}={{e}^{\dfrac{1}{x}}}\left( -\dfrac{1}{{{x}^{2}}} \right)=-\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}$
We are asked to differentiate the function $y=\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}$, we are asked to find its derivative. We have to use the quotient rule to solve this problem.
The quotient rule states that $\dfrac{d\left( \dfrac{f(x)}{g(x)} \right)}{dx}=\dfrac{\dfrac{d\left( f(x) \right)}{dx}g(x)-f(x)\dfrac{d\left( g(x) \right)}{dx}}{{{\left( g(x) \right)}^{2}}}$. Here, we have $f(x)={{e}^{\dfrac{1}{x}}}\And g(x)={{x}^{2}}$. Hence, using the quotient rule we can differentiate it as follows
$\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}} \right)}{dx}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{d\left( {{e}^{\dfrac{1}{x}}} \right)}{dx}{{x}^{2}}-{{e}^{\dfrac{1}{x}}}\dfrac{d\left( {{x}^{2}} \right)}{dx}}{{{\left( {{x}^{2}} \right)}^{2}}}$
We have already evaluated the derivative of ${{e}^{\dfrac{1}{x}}}$ using the composite function method, and we know the derivative of ${{x}^{2}}$ with respect to x is $2x$. Substituting, these expression in the above differentiation, we get the derivative as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\dfrac{{{e}^{\dfrac{1}{x}}}}{{{x}^{2}}}\times {{x}^{2}}-{{e}^{\dfrac{1}{x}}}\left( 2x \right)}{{{\left( {{x}^{2}} \right)}^{2}}}$
Simplifying the above expression, we get the derivative as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-{{e}^{\dfrac{1}{x}}}-2x{{e}^{\dfrac{1}{x}}}}{{{x}^{4}}}$
**We can take the term ${{e}^{\dfrac{1}{x}}}$ as a common factor from the terms in the numerators, to express the above expression as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-\left( 1+2x \right){{e}^{\dfrac{1}{x}}}}{{{x}^{4}}}$ **

Note: One must know the derivatives of different functions to solve these types of problems. Along with them, we should also know the different properties like product rule and quotient rule for differentiating complex functions.