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

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

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

Answer

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})