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Question: How do you find f’(x) using the definition of a derivative \(y=6{{e}^{x}}+\dfrac{4}{\sqrt[3]{x}}\)?...

How do you find f’(x) using the definition of a derivative y=6ex+4x3y=6{{e}^{x}}+\dfrac{4}{\sqrt[3]{x}}?

Explanation

Solution

Assume the given function as f(x). Then it’s derivative will be f’(x). Apply the sum rule of derivative to separate the functions. Take out the constants and apply the power rule of derivative to find the derivative of the functions. Write the different derivative values together to obtain the value of f’(x).

Complete step by step solution:
Given y= f(x)=6ex+4x36{{e}^{x}}+\dfrac{4}{\sqrt[3]{x}}
We know the derivative of f(x) is f’(x)
f′(x)=ddxf(x) ⇒f′(x)=ddx(6ex+4x3) \begin{aligned} & f'\left( x \right)=\dfrac{d}{dx}f\left( x \right) \\\ & \Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( 6{{e}^{x}}+\dfrac{4}{\sqrt[3]{x}} \right) \\\ \end{aligned}
Applying sum rule of derivative, we get
⇒f′(x)=ddx(6ex)+ddx(4x3)\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( 6{{e}^{x}} \right)+\dfrac{d}{dx}\left( \dfrac{4}{\sqrt[3]{x}} \right)
As we know, the derivative of ex{{e}^{x}}is ex{{e}^{x}}.
So, ddx(6ex)\dfrac{d}{dx}\left( 6{{e}^{x}} \right) can be written as ddx(6ex)=6ddx(ex)=6ex\dfrac{d}{dx}\left( 6{{e}^{x}} \right)=6\dfrac{d}{dx}\left( {{e}^{x}} \right)=6{{e}^{x}} (taking out constant (a⋅f)′=a⋅f′\left( a\cdot f \right)'=a\cdot f')
4x3\dfrac{4}{\sqrt[3]{x}}can be written as 4x134{{x}^{\dfrac{1}{3}}}
Again as we know the power rule of derivative is xn=nxn−1{{x}^{n}}=n{{x}^{n-1}}
So, ddx(4x3)\dfrac{d}{dx}\left( \dfrac{4}{\sqrt[3]{x}} \right) can be written as ddx(4x3)=ddx(4x13)=4×(−13)x−13−1=−43x−1−33=−43x−43\dfrac{d}{dx}\left( \dfrac{4}{\sqrt[3]{x}} \right)=\dfrac{d}{dx}\left( 4{{x}^{\dfrac{1}{3}}} \right)=4\times \left( -\dfrac{1}{3} \right){{x}^{-\dfrac{1}{3}-1}}=-\dfrac{4}{3}{{x}^{\dfrac{-1-3}{3}}}=-\dfrac{4}{3}{{x}^{-\dfrac{4}{3}}}
Now, f’(x) becomes
⇒f′(x)=6ex+(−43x−43)\Rightarrow f'\left( x \right)=6{{e}^{x}}+\left( -\dfrac{4}{3}{{x}^{-\dfrac{4}{3}}} \right)
Again 4x−434{{x}^{-\dfrac{4}{3}}} can be written as 4x43\dfrac{4}{\sqrt[3]{{{x}^{4}}}}
So, f’(x) becomes
⇒f′(x)=6ex−4x43\Rightarrow f'\left( x \right)=6{{e}^{x}}-\dfrac{4}{\sqrt[3]{{{x}^{4}}}}
This is the required solution of the given question.

Note:
According to the definition of derivative the derivative of f(x) with respect to ‘x’ is the function f’(x) and is defined as f′(x)=lim⁡f(x+h)−f(x)hf'\left( x \right)=\lim \dfrac{f\left( x+h \right)-f\left( x \right)}{h}. The above result can be verified using this formula.