Question Video: Finding the First Derivative of a Function Defined by Parametric Equations Involving Using the Product Rule | Nagwa Question Video: Finding the First Derivative of a Function Defined by Parametric Equations Involving Using the Product Rule | Nagwa

# Question Video: Finding the First Derivative of a Function Defined by Parametric Equations Involving Using the Product Rule Mathematics • Third Year of Secondary School

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Given that π₯ = 3π^(5π‘) and π¦ = π‘π^(β5π‘), find dπ¦/dπ₯.

03:08

### Video Transcript

Given that π₯ equals three π to the power of five π‘ and π¦ equals π‘ times π to the power of negative five π‘, find dπ¦ by dπ₯.

In this question, weβve been given a pair of parametric equations. These are equations for π₯ and π¦ in terms of a third parameter; here, thatβs π‘. Weβre asked to differentiate π¦ with respect to π₯. And so we recall that if dπ¦ by dπ‘ is not equal to zero, dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ divided by dπ₯ by dπ‘. And so we can see that weβre going to need to differentiate the equation for π₯ with respect to π‘ and the equation for π¦ with respect to π‘.

Weβll begin by differentiating π₯ with respect to π‘. Now, π₯ is equal to three π to the power of five π‘. And we know that to differentiate π to the power of ππ₯ for some constant value π, we get π times π to the power of ππ₯. Well, in this case, weβre differentiating three π to the power of five π‘. So we multiply the entire term by the coefficient of π‘, which is five. And we see that dπ₯ by dπ‘ is five times three π to the power of five π‘, which is 15π to the power of five π‘.

Next, we need to differentiate π¦ with respect to π‘. But we see that π¦ itself is the product of two differentiable functions. Itβs π‘ times π to the power of negative five π‘. And so weβre going to use the product rule. This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. Well, our two functions are π‘ and π to the power of negative five π‘. So dπ¦ by dπ‘ is π‘ times the derivative of π to the power of negative five π‘ with respect to π‘ plus π to the power of negative five π‘ times the derivative of π‘ with respect to π‘.

Now, the derivative of π to the power of negative five π‘ we know to be equal to negative five π to the power of negative five π‘. And the derivative of π‘ with respect to π‘ is simply one. So dπ¦ by dπ‘ is π‘ times negative five π to the power of negative five π‘ plus π to the power of negative five π‘, which simplifies as shown. dπ¦ by dπ₯ is the quotient of these. Itβs dπ¦ by dπ‘ divided by dπ₯ by dπ‘. So thatβs negative five π‘π to the power of negative five π‘ plus π to the power of negative five π‘ all over 15π to the power five π‘.

By recognizing that π to the power of negative five π‘ is the same as one over π to the power of five π‘ and then factoring by one over π to the power of five π‘. We see we can write this as one over π to the power of five π‘ times negative five π‘ plus one over 15π to the power of five π‘. But of course, we would generally write this second fraction as one minus five π‘ over 15π to the power of five π‘. Now, when we multiply two numbers with exponents, we add those exponents. So π to the power of five π‘ times 15π to the power of five π‘ is 15π to the power of 10π‘.

And we therefore find dπ¦ by dπ₯ to be equal to one minus five π‘ over 15π to the power of 10π‘.

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