Question Video: Differentiating Parametric Functions Involving Trigonometric Ratios and Natural Exponents | Nagwa Question Video: Differentiating Parametric Functions Involving Trigonometric Ratios and Natural Exponents | Nagwa

# Question Video: Differentiating Parametric Functions Involving Trigonometric Ratios and Natural Exponents Mathematics • Third Year of Secondary School

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Given that π₯ = 5π‘π^(π‘) and π¦ = 3π‘ + 4 sin π‘, find dπ¦/dπ₯.

03:32

### Video Transcript

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

In this case, instead of π¦ being defined as a function of π₯, both π₯ and π¦ have been defined in terms of a third variable π‘. And we call these parametric equations. To find dπ¦ by dπ₯, we remind ourselves of the chain rule, which gives us the equation dπ¦ by dπ‘ is equal to dπ¦ by dπ₯ multiplied by dπ₯ by dπ‘. We can rearrange this for dπ¦ by dπ₯. And we get that dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘. And of course, this only works providing that dπ₯ by dπ‘, the denominator, is not equal to zero.

So, weβre going to need to find dπ¦ by dπ‘ and dπ₯ by dπ‘. dπ¦ by dπ‘ is the derivative of π¦ equals three π‘ add four sin of π‘ with respect to π‘. We can do this term by term, starting with the term three π‘, which we know differentiates to give us three. Now for the term four sin of π‘, we recall that the derivative of sin of π‘ is cos of π‘ with respect to π‘. So, four sin of π‘ differentiates to give us four cos of π‘. So we have that dπ¦ by dπ‘ is equal to three add four cos of π‘.

Now weβve also got to find dπ₯ by dπ‘. This is going to be the derivative of π₯ equals five π‘π to the π‘ power with respect to π‘. One thing to be aware of here is that we have two functions of π‘ being multiplied together. Itβs five π‘ multiplied by π to the π‘ power. So, to differentiate this, weβve got to use the product rule. The product rule tells us that the derivative of function π multiplied by function π is π prime multiplied by π add π multiplied by π prime. Remember, the notation π prime just means the derivative of the function π.

In this case, the function π of π‘ is equal to five π‘ and the function π of π‘ is equal to π to the π‘ power. So in order to use the product rule, we want to find π prime of π‘ and π prime of π‘, the derivative of π and the derivative of π. Well, we know that the derivative of five π‘ is five. To differentiate the function π, we recall the fact that the derivative of π to the π‘ power is just π to the π‘ power. And this is just a general result of differentiation of the exponential function.

Now we can put what we found into the formula for the product rule. So we find that dπ₯ by dπ‘ is equal to π prime multiplied by π, which is five multiplied by π to the π‘ power add π multiplied by π prime, which is five π‘ multiplied by π to the π‘ power. So by the product rule, dπ₯ by dπ‘ is five π to the π‘ power add five π‘π to the π‘ power. Notice that we can take five π to the π‘ power out as a common factor, which gives us five π to the π‘ power multiplied by one add π‘. Or equivalently, we could write this the other way as five π to the π‘ power multiplied by π‘ add one.

We can now use the fact that dπ¦ by dπ₯ equals dπ¦ by dπ‘ over dπ₯ by dπ‘. Then we have that dπ¦ by dπ₯ equals three add four cos of π‘ over five π to the π‘ multiplied by π‘ add one. And that gives us our final answer.

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