Question

31: let f(x) be a differentiable function. suppose f(0)=5 and f(0)=\frac{1}{5}. consider g(x)=f(x)\sin x. which of the following is equal to the slope of the line tangent to the graph of g(x) at x = 0? (a) - 4 (b) 0 (c) \frac{1}{5} (d) 5 (e) it does not exist at x = 0.

Explanation:

Step1: Apply product - rule for differentiation

The product - rule states that if $g(x)=u(x)v(x)$, then $g^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x) = f(x)$ and $v(x)=\sin x$. So, $g^{\prime}(x)=f^{\prime}(x)\sin x + f(x)\cos x$.

Step2: Evaluate $g^{\prime}(x)$ at $x = 0$

Substitute $x = 0$ into $g^{\prime}(x)$. We know that $f(0)=5$ and $f^{\prime}(0)=\frac{1}{5}$, and $\sin(0)=0$, $\cos(0)=1$. Then $g^{\prime}(0)=f^{\prime}(0)\sin(0)+f(0)\cos(0)$.

Step3: Calculate the value of $g^{\prime}(0)$

Since $f^{\prime}(0)\sin(0)=\frac{1}{5}\times0 = 0$ and $f(0)\cos(0)=5\times1 = 5$, then $g^{\prime}(0)=0 + 5=5$.

Answer:

D. 5