Question

find $g(7)$ if $g(x)$ is a differentiable function satisfying
$g(x)+x\sin(g(x)) = x^{2}-49$ and $g(7)=0$.
$g(7)= $

Explanation:

Step1: Differentiate both sides

Differentiate \(g(x)+x\sin(g(x)) = x^{2}-49\) with respect to \(x\) using sum rule \((u + v)^\prime=u^\prime+v^\prime\), product rule \((uv)^\prime = u^\prime v+uv^\prime\) and chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\).

For the left - hand side:
The derivative of \(g(x)\) is \(g^\prime(x)\).
For \(x\sin(g(x))\), let \(u = x\) and \(v=\sin(g(x))\). Then \(u^\prime=1\) and \(v^\prime=\cos(g(x))\cdot g^\prime(x)\). So \((x\sin(g(x)))^\prime=\sin(g(x))+x\cos(g(x))g^\prime(x)\).

The derivative of the right - hand side: \((x^{2}-49)^\prime = 2x\).

So, \(g^\prime(x)+\sin(g(x))+x\cos(g(x))g^\prime(x)=2x\).

Step2: Substitute \(x = 7\)

Since \(g(7) = 0\), substitute \(x = 7\) and \(g(7)=0\) into \(g^\prime(x)+\sin(g(x))+x\cos(g(x))g^\prime(x)=2x\).

We get \(g^\prime(7)+\sin(0)+7\cos(0)g^\prime(7)=2\times7\).

Because \(\sin(0)=0\) and \(\cos(0) = 1\), the equation becomes \(g^\prime(7)+0 + 7\times1\times g^\prime(7)=14\).

Combine like terms: \((1 + 7)g^\prime(7)=14\), i.e., \(8g^\prime(7)=14\).

Step3: Solve for \(g^\prime(7)\)

Divide both sides of \(8g^\prime(7)=14\) by \(8\). So \(g^\prime(7)=\frac{14}{8}=\frac{7}{4}\).

Answer:

\(\frac{7}{4}\)