Video: Differentiating Parametric Functions Involving Trigonometric Ratios and Natural Exponents

Given that π‘₯ = 5𝑑𝑒^(𝑑) and 𝑦 = 3𝑑 + 4 sin 𝑑, find d𝑦/dπ‘₯.

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Video Transcript

Given that π‘₯ equals five 𝑑𝑒 to the 𝑑 power and 𝑦 equals three 𝑑 plus four sin of 𝑑, find d𝑦 by dπ‘₯.

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

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

Now we’ve also got to find dπ‘₯ by d𝑑. This is going to be the derivative of π‘₯ equals five 𝑑𝑒 to the 𝑑 power with respect to 𝑑. One thing to be aware of here is that we have two functions of 𝑑 being multiplied together. It’s five 𝑑 multiplied by 𝑒 to the 𝑑 power. So, to differentiate this, we’ve got to use the product rule. The product rule tells us that the derivative of function 𝑓 multiplied by function 𝑔 is 𝑓 prime multiplied by 𝑔 add 𝑓 multiplied by 𝑔 prime. Remember, the notation 𝑓 prime just means the derivative of the function 𝑓.

In this case, the function 𝑓 of 𝑑 is equal to five 𝑑 and the function 𝑔 of 𝑑 is equal to 𝑒 to the 𝑑 power. So in order to use the product rule, we want to find 𝑓 prime of 𝑑 and 𝑔 prime of 𝑑, the derivative of 𝑓 and the derivative of 𝑔. Well, we know that the derivative of five 𝑑 is five. To differentiate the function 𝑔, we recall the fact that the derivative of 𝑒 to the 𝑑 power is just 𝑒 to the 𝑑 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 d𝑑 is equal to 𝑓 prime multiplied by 𝑔, which is five multiplied by 𝑒 to the 𝑑 power add 𝑓 multiplied by 𝑔 prime, which is five 𝑑 multiplied by 𝑒 to the 𝑑 power. So by the product rule, dπ‘₯ by d𝑑 is five 𝑒 to the 𝑑 power add five 𝑑𝑒 to the 𝑑 power. Notice that we can take five 𝑒 to the 𝑑 power out as a common factor, which gives us five 𝑒 to the 𝑑 power multiplied by one add 𝑑. Or equivalently, we could write this the other way as five 𝑒 to the 𝑑 power multiplied by 𝑑 add one.

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

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