Question

assume that $x = x(t)$ and $y = y(t)$. let $y = x^{2}+5$ and $\frac{dx}{dt}=4$ when $x = 1$.
find $\frac{dy}{dt}$ when $x = 1$.
$\frac{dy}{dt}=square$ (simplify your answer.)

Explanation:

Step1: Differentiate \(y\) with respect to \(t\)

Using the chain rule, if \(y = x^{2}+5\), then \(\frac{dy}{dt}=\frac{d}{dt}(x^{2}+5)\).
By the sum rule \(\frac{d}{dt}(u + v)=\frac{du}{dt}+\frac{dv}{dt}\) (here \(u = x^{2}\), \(v = 5\)) and the chain - rule \(\frac{d}{dt}(x^{n})=nx^{n - 1}\frac{dx}{dt}\), we have \(\frac{dy}{dt}=2x\frac{dx}{dt}+0\).

Step2: Substitute the given values

We are given that \(x = 1\) and \(\frac{dx}{dt}=4\).
Substitute \(x = 1\) and \(\frac{dx}{dt}=4\) into the equation \(\frac{dy}{dt}=2x\frac{dx}{dt}\).
So \(\frac{dy}{dt}=2\times1\times4\).

Answer:

\(8\)