Question

math 100 w25 t1 assignment 5
topics. chain rule, implicit differentiation, lhôpitals rule, related rates

1. suppose that ( f(x) ) is a function that satisfies ( f(4)=5 ) and ( f^{prime}(4)=-3 ). let ( g(x)=e^{-x^{2}}+ln (f(x)) ). calculate ( g^{prime}(4) ).

Explanation:

Step1: Differentiate \(g(x)\) using sum rule and chain rule

The sum rule states \((u + v)^\prime=u^\prime + v^\prime\). Let \(u = e^{-x^{2}}\) and \(v=\ln(f(x))\).
For \(u = e^{-x^{2}}\), by the chain rule \((e^{h(x)})^\prime=e^{h(x)}\cdot h^\prime(x)\), where \(h(x)=-x^{2}\), so \(u^\prime=e^{-x^{2}}\cdot(-2x)\).
For \(v = \ln(f(x))\), by the chain rule \((\ln(k(x)))^\prime=\frac{k^\prime(x)}{k(x)}\), where \(k(x)=f(x)\), so \(v^\prime=\frac{f^\prime(x)}{f(x)}\).
Then \(g^\prime(x)=-2x e^{-x^{2}}+\frac{f^\prime(x)}{f(x)}\).

Step2: Substitute \(x = 4\) into \(g^\prime(x)\)

When \(x = 4\), we know that \(f(4) = 5\) and \(f^\prime(4)=-3\).
Substitute into \(g^\prime(x)\):
\(g^\prime(4)=-2\times4\times e^{-4^{2}}+\frac{-3}{5}\).
\(g^\prime(4)=-8e^{-16}-\frac{3}{5}\).

Answer:

\(g^\prime(4)=-8e^{-16}-\frac{3}{5}\)