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Question: How do you find the derivative of \(f\left( x \right)={{\left( 2-4{{e}^{2x}} \right)}^{3}}\) ?...

How do you find the derivative of f(x)=(2−4e2x)3f\left( x \right)={{\left( 2-4{{e}^{2x}} \right)}^{3}} ?

Explanation

Solution

We can find the derivative of f(x)=(2−4e2x)3f\left( x \right)={{\left( 2-4{{e}^{2x}} \right)}^{3}} with respect to x by chain rule, the chain rule is if we have find the derivative of f(g(x)) we can assume g(x) as t and differentiate f with respect to t and then differentiate t with respect to x.

Complete step by step answer:
The given equation which we have to differentiate with respect to x is f(x)=(2−4e2x)3f\left( x \right)={{\left( 2-4{{e}^{2x}} \right)}^{3}}
We know that we can write dydx=dydt×dtdx\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}
So let’s assume (2−4e2x)\left( 2-4{{e}^{2x}} \right) as t so f(x) is equal to t3{{t}^{3}}
So we can write dfdx=dfdt×dtdx\dfrac{df}{dx}=\dfrac{df}{dt}\times \dfrac{dt}{dx}
⇒dfdx=dt3dt×dtdx\Rightarrow \dfrac{df}{dx}=\dfrac{d{{t}^{3}}}{dt}\times \dfrac{dt}{dx}
Derivative of t3{{t}^{3}} with respect to t is 3t23{{t}^{2}}, we know that derivative of xn{{x}^{n}} is nxn−1n{{x}^{n-1}} where n is not equal to 0.
So we can write
⇒dfdx=t3×dtdx\Rightarrow \dfrac{df}{dx}={{t}^{3}}\times \dfrac{dt}{dx}
We can replace t with (2−4e2x)\left( 2-4{{e}^{2x}} \right)
⇒dfdx=3(2−4e2x)2×d(2−4e2x)dx\Rightarrow \dfrac{df}{dx}=3{{\left( 2-4{{e}^{2x}} \right)}^{2}}\times \dfrac{d\left( 2-4{{e}^{2x}} \right)}{dx}
Let’s find the derivative of (2−4e2x)\left( 2-4{{e}^{2x}} \right) with respect to x. We know the derivative of a constant term is 0. So the derivative of 2 is 0. Derivative of e2x{{e}^{2x}} is equal to 2e2x2{{e}^{2x}} so derivative of (2−4e2x)\left( 2-4{{e}^{2x}} \right) is equal to −8e2x-8{{e}^{2x}}
So the value of d(2−4e2x)dx\dfrac{d\left( 2-4{{e}^{2x}} \right)}{dx} is equal to −8e2x-8{{e}^{2x}} we can replace it
⇒dfdx=3(2−4e2x)2×−8e2x\Rightarrow \dfrac{df}{dx}=3{{\left( 2-4{{e}^{2x}} \right)}^{2}}\times -8{{e}^{2x}}
⇒dfdx=−24(2−4e2x)2e2x\Rightarrow \dfrac{df}{dx}=-24{{\left( 2-4{{e}^{2x}} \right)}^{2}}{{e}^{2x}}

Note:
Always remember that the derivative of xn{{x}^{n}} is nxn−1n{{x}^{n-1}} where n is not equal to 0 because when n is equal to 0 the value of xn{{x}^{n}} becomes a constant term and we know that the derivative of a constant is always 0 . The integration of xn{{x}^{n}} is equal to xn+1n+1\dfrac{{{x}^{n+1}}}{n+1} where n is not equal to – 1, if n is equal to – 1 then xn{{x}^{n}} will be 1x\dfrac{1}{x} and if we put n= -1 in the formula the denominator will be 0 , integration of 1x\dfrac{1}{x} is lnx .