Question

a) $g(x)=(x^{2}+2)^{4}(x^{2}+5)^{7}$ $g(x)=(x^{2}+3)^{4}cdot(x^{2}+8)^{7}$ $=(x^{2}+3)^{4}cdot(x^{2}+5)^{7}+(x^{2}+3)^{4}cdot(x^{2}+5)^{7}$ $=4(x^{2}+3)^{4 - 1}cdot(x^{2}+3)cdot(x^{2}+5)^{7}+(x^{2}+3)^{4}cdot7(x^{2}+5)^{7 - 1}cdot(x^{2}+5)$ $4(x^{2}+3)^{3}cdot(2x)cdot(x^{2}+5)^{7}+(x^{2}+3)^{4}cdot7(x^{2}+5)^{6}cdot(2x)$ b) $y=sqrt{1+xe^{-3x}}$ $y=(sqrt{1+xe^{-3x}})=(1+xe^{-3x})^{1/2}=$ $\frac{1}{2}(1+xe^{-3x})^{1/2 - 1}cdot(1+xe^{-2x})=\frac{1}{2}(1+xe^{-3x})^{-1/2}(xe^{-2})$ $=\frac{1}{2}\frac{1}{(1+xe^{-3x})^{1/2}}cdot((x)e^{-3x}+x(e^{-3x}))$ $=\frac{1}{2}\frac{1}{sqrt{1+xe^{-3x}}}cdot(e^{-3x}+xcdot e^{-3x}cdot(-3))$

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(y'=u'v + uv'\). For \(g(x)=(x^{2}+3)^{4}(x^{2}+5)^{7}\), let \(u=(x^{2}+3)^{4}\) and \(v=(x^{2}+5)^{7}\).
First, find \(u'\) using the chain rule. If \(u = (x^{2}+3)^{4}\), let \(t=x^{2}+3\), then \(u = t^{4}\). By the chain rule \(\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}\). \(\frac{du}{dt} = 4t^{3}\) and \(\frac{dt}{dx}=2x\), so \(u'=4(x^{2}+3)^{3}\cdot2x = 8x(x^{2}+3)^{3}\).
Second, find \(v'\) using the chain rule. If \(v=(x^{2}+5)^{7}\), let \(s=x^{2}+5\), then \(v = s^{7}\). By the chain rule \(\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}\). \(\frac{dv}{ds}=7s^{6}\) and \(\frac{ds}{dx}=2x\), so \(v'=14x(x^{2}+5)^{6}\).
Then \(g'(x)=u'v+uv'=8x(x^{2}+3)^{3}(x^{2}+5)^{7}+14x(x^{2}+3)^{4}(x^{2}+5)^{6}\).
Factor out \(2x(x^{2}+3)^{3}(x^{2}+5)^{6}\):
\(g'(x)=2x(x^{2}+3)^{3}(x^{2}+5)^{6}[4(x^{2}+5)+7(x^{2}+3)]\)
\(=2x(x^{2}+3)^{3}(x^{2}+5)^{6}(4x^{2}+20 + 7x^{2}+21)\)
\(=2x(x^{2}+3)^{3}(x^{2}+5)^{6}(11x^{2}+41)\)

Step2: Apply the chain rule for \(y = \sqrt{1+xe^{-3x}}=(1+xe^{-3x})^{\frac{1}{2}}\)

Let \(u = 1+xe^{-3x}\), then \(y = u^{\frac{1}{2}}\). By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}\)
Now find \(\frac{du}{dx}\). Using the sum rule \(\frac{du}{dx}=\frac{d(1)}{dx}+\frac{d(xe^{-3x})}{dx}\). Since \(\frac{d(1)}{dx} = 0\), for \(\frac{d(xe^{-3x})}{dx}\), use the product rule. Let \(m = x\) and \(n=e^{-3x}\). Then \(\frac{d(xe^{-3x})}{dx}=m'n+mn'\). \(m' = 1\), \(n'=- 3e^{-3x}\) (by the chain rule, if \(n = e^{-3x}\), let \(p=-3x\), \(n = e^{p}\), \(\frac{dn}{dx}=\frac{dn}{dp}\cdot\frac{dp}{dx}=e^{p}\cdot(-3)\)). So \(\frac{d(xe^{-3x})}{dx}=e^{-3x}-3xe^{-3x}\)
Then \(\frac{dy}{dx}=\frac{1}{2}(1+xe^{-3x})^{-\frac{1}{2}}(e^{-3x}-3xe^{-3x})=\frac{e^{-3x}(1 - 3x)}{2\sqrt{1+xe^{-3x}}}\)

Answer:

a) \(g'(x)=2x(x^{2}+3)^{3}(x^{2}+5)^{6}(11x^{2}+41)\)
b) \(y'=\frac{e^{-3x}(1 - 3x)}{2\sqrt{1+xe^{-3x}}}\)