Question

if ( y = (x^{2}+9)^{4} ), find ( \frac{d^{2}y}{dx^{2}} )
( \frac{d^{2}y}{dx^{2}}=square )

Explanation:

Step1: Find the first - derivative using the chain rule

The chain rule states that if \(y = u^n\) where \(u\) is a function of \(x\), then \(\frac{dy}{dx}=n\cdot u^{n - 1}\cdot\frac{du}{dx}\).
Let \(u=x^{2}+9\) and \(n = 4\). Then \(\frac{du}{dx}=2x\).
So, \(\frac{dy}{dx}=4(x^{2}+9)^{3}\cdot(2x)=8x(x^{2}+9)^{3}\).

Step2: Find the second - derivative using the product rule

The product rule states that if \(y = f(x)\cdot g(x)\), then \(\frac{dy}{dx}=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)\).
Let \(f(x)=8x\) and \(g(x)=(x^{2}+9)^{3}\).
First, find \(f^{\prime}(x)\) and \(g^{\prime}(x)\):
\(f^{\prime}(x)=8\).
For \(g^{\prime}(x)\), use the chain rule again. Let \(v=x^{2}+9\), \(m = 3\). Then \(\frac{dv}{dx}=2x\), and \(g^{\prime}(x)=3(x^{2}+9)^{2}\cdot(2x)=6x(x^{2}+9)^{2}\).
Now, by the product rule:
\(\frac{d^{2}y}{dx^{2}}=8\cdot(x^{2}+9)^{3}+8x\cdot6x(x^{2}+9)^{2}\).
Factor out \(8(x^{2}+9)^{2}\):
\(\frac{d^{2}y}{dx^{2}}=8(x^{2}+9)^{2}[(x^{2}+9)+6x^{2}]\).
Simplify the expression inside the brackets: \((x^{2}+9)+6x^{2}=7x^{2}+9\).

Answer:

\(8(x^{2}+9)^{2}(7x^{2}+9)\)