# If y = 5x^3 - 10x, determine [dy/dt] when x = 0 and [dx/dt] = -2 .

## Question:

If {eq}y = 5x^3 - 10x {/eq}, determine {eq}\frac{dy}{dt} {/eq}, when x = 0 and {eq}\frac{dx}{dt} {/eq}= -2.

## Parametric Derivative:

In parametric derivative, value of {eq}\frac{dy}{dt} {/eq} and {eq}\frac{dx}{dt} {/eq} is use to derive value of {eq}\frac{dx}{dy} {/eq}.

In this question we need to differentiate whole equation with respect to t. And so after putting given values, we will get to obtain the required result.

## Answer and Explanation:

We have,

{eq}y = 5x^3 - 10x {/eq}

Now,

Differentiate with respect to t both sides,

{eq}\frac{dy}{dt} = (5x^3 - 10x)' \\ \frac{dy}{dt} = 15x^2 \frac{dx}{dt} - 10 \frac{dx}{dt} \\ {/eq}

On replacing the values,

{eq}\frac{dx}{dt} {/eq}= -2 and x = 0

so,

{eq}\frac{dy}{dt} = 15(0)^2 (-2) - 10 (-2) \\ \frac{dy}{dt} = 20 \\ {/eq}

{eq}\therefore \color{blue}{\frac{dy}{dt} = 20} {/eq}

#### Learn more about this topic:

Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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