Question

differentiate the following function.

f(x)=e^{5left(x^{2}+4
ight)^{2}}

f^{prime}(x)=20 xleft(x^{2}+4
ight) e^{5left(x^{2}+4
ight)^{2}}

Explanation:

Step1: Let \(u = 5(x^{2}+4)^{2}\)

The function \(y = f(x)=e^{u}\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). First, find \(\frac{dy}{du}\).
Since \(y = e^{u}\), then \(\frac{dy}{du}=e^{u}\).

Step2: Find \(\frac{du}{dx}\)

Let \(v=(x^{2}+4)\), so \(u = 5v^{2}\).
First, \(\frac{du}{dv}=10v\) (using the power rule \(\frac{d}{dv}(av^{n})=anv^{n - 1}\), here \(a = 5\), \(n = 2\)).
Then \(\frac{dv}{dx}=2x\) (using the power rule \(\frac{d}{dx}(x^{n})=nx^{n-1}\), here \(n = 2\) for \(x^{2}\) and \(\frac{d}{dx}(4)=0\)).
By the chain - rule \(\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}\). Substitute \(v=(x^{2}+4)\):
\(\frac{du}{dx}=10(x^{2}+4)\cdot2x=20x(x^{2}+4)\).

Step3: Combine using the chain - rule

Since \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\) and \(u = 5(x^{2}+4)^{2}\), \(\frac{dy}{du}=e^{u}\), \(\frac{du}{dx}=20x(x^{2}+4)\)
\(f^{\prime}(x)=e^{5(x^{2}+4)^{2}}\cdot20x(x^{2}+4)\)

Answer:

\(f^{\prime}(x)=20x(x^{2}+4)e^{5(x^{2}+4)^{2}}\)