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Question: How do you differentiate \(f(x)={{(6{{e}^{-x}}+2)}^{3}}\) using the chain rule?...

How do you differentiate f(x)=(6e−x+2)3f(x)={{(6{{e}^{-x}}+2)}^{3}} using the chain rule?

Explanation

Solution

The expression given is to be differentiated using the chain rule. Firstly, we will differentiate the power and we will get 3(6e−x+2)23{{(6{{e}^{-x}}+2)}^{2}}. Then, we will proceed on to differentiate the (6e−x+2)(6{{e}^{-x}}+2). Differentiation of 6e−x6{{e}^{-x}} gives us the value −6e−x-6{{e}^{-x}} and derivative of 2 is 0. Hence, we have the differentiation of the given function.

Complete step-by-step solution:
According to the given question, we have been given an expression which we have to differentiate using the chain rule.
And as we know using chain rule implies that we will have to begin the differentiation from the outermost function and proceed to the innermost function.
The expression we have is,
f(x)=(6e−x+2)3f(x)={{(6{{e}^{-x}}+2)}^{3}}-----(1)
We will first differentiate the power of the given function, and so we get,
f′(x)=3(6e−x+2)2f'(x)=3{{(6{{e}^{-x}}+2)}^{2}}------(2)
We have the above expression as we know, if f(x)=x3f(x)={{x}^{3}} then f′(x)=3x2f'(x)=3{{x}^{2}}.
We continue with the differentiation, we will now differentiate the (6e−x+2)(6{{e}^{-x}}+2) term, that is,
f′(x)=3(6e−x+2)2.(ddx(6e−x+2))f'(x)=3{{(6{{e}^{-x}}+2)}^{2}}.\left( \dfrac{d}{dx}(6{{e}^{-x}}+2) \right)-----(3)
On solving further, we get the expression as,
f′(x)=3(6e−x+2)2.(ddx(6e−x)+ddx(2))f'(x)=3{{(6{{e}^{-x}}+2)}^{2}}.\left( \dfrac{d}{dx}(6{{e}^{-x}})+\dfrac{d}{dx}(2) \right)
We know that the derivative of 6e−x6{{e}^{-x}} is −6e−x-6{{e}^{-x}}.
As ddx(6e−x)=6e−x.(−1)=−6e−x\dfrac{d}{dx}(6{{e}^{-x}})=6{{e}^{-x}}.(-1)=-6{{e}^{-x}} and we also know that the derivative of a constant like 2 is 0.
And so we get the expression as,
f′(x)=3(6e−x+2)2.(6ddx(e−x)+0)f'(x)=3{{(6{{e}^{-x}}+2)}^{2}}.\left( 6\dfrac{d}{dx}({{e}^{-x}})+0 \right)
f′(x)=3(6e−x+2)2.(−6e−x)f'(x)=3{{(6{{e}^{-x}}+2)}^{2}}.\left( -6{{e}^{-x}} \right)
f′(x)=−18(6e−x+2)2.(e−x)f'(x)=-18{{(6{{e}^{-x}}+2)}^{2}}.\left( {{e}^{-x}} \right)
Therefore, the differentiation of the given expression is:
f′(x)=−18(6e−x+2)2.(e−x)f'(x)=-18{{(6{{e}^{-x}}+2)}^{2}}.\left( {{e}^{-x}} \right).

Note: While differentiating e−x{{e}^{-x}}, care should be taken that after the exponential function is differentiated, the −x-x term in the power is also differentiated. This is also according to the chain rule. So, we have the derivative of e−x{{e}^{-x}} as:
ddx(e−x)=e−x.(−1)=−e−x\dfrac{d}{dx}\left( {{e}^{-x}} \right)={{e}^{-x}}.(-1)=-{{e}^{-x}}