# Assume that {eq}x=x(t) {/eq} and {eq}y=y(t) {/eq}. Let {eq}y=x^3+1 {/eq} and {eq}\displaystyle \frac{dx}{dt} = 2 \ {/eq} when {eq}x = 1 {/eq}. Find {eq}\displaystyle \frac{dy}{dt} {/eq} when {eq}x = 1 {/eq}.

## Question:

Assume that {eq}x=x(t) {/eq} and {eq}y=y(t) {/eq}. Let {eq}y=x^3+1 {/eq} and {eq}\displaystyle \frac{dx}{dt} = 2 \ {/eq} when {eq}x = 1 {/eq}.

Find {eq}\displaystyle \frac{dy}{dt} {/eq} when {eq}x = 1 {/eq}.

## Chain Rule for Derivatives:

We are given that y is a function of x, and x is a function of t, which means y is also a function of t. So, we find {eq}\displaystyle \frac{\mathrm{d} y(t)}{\mathrm{d} t} {/eq} using the chain rule. Now, the values of x and {eq}\displaystyle \frac{\mathrm{d} x}{\mathrm{d} t} {/eq} are substituted to get the value of {eq}\displaystyle \frac{\mathrm{d} y}{\mathrm{d} t}. {/eq}

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{eq}y=x^3+1 {/eq}

Differentiate with respect to t, we get

{eq}\displaystyle \frac{dy}{dt}=3x^2 \frac{dx}{dt} {/eq}

Given {eq}\displaystyle... 