Question Video: Differentiating a Combination of Trigonometric and Linear Functions at a Point Mathematics • Higher Education

Find the derivative of 7π‘₯ + 4 sin π‘₯ with respect to cos π‘₯ + 1 at π‘₯ = πœ‹/6.


Video Transcript

Find the derivative of seven π‘₯ plus four sin π‘₯ with respect to cos π‘₯ plus one at π‘₯ equals πœ‹ by six.

Let’s begin by defining our two functions. We’ll let 𝑦 be equal to seven π‘₯ plus four sin π‘₯. And we’ll define cos π‘₯ plus one as 𝑧. We then recall that, given two parametric equations β€” π‘₯ equals 𝑓 of 𝑑 and 𝑦 equals 𝑔 of 𝑑 β€” we can find d𝑦 by dπ‘₯ by multiplying d𝑦 by d𝑑 by one over dπ‘₯ by d𝑑. Or equivalently, by dividing d𝑦 by d𝑑 by dπ‘₯ by d𝑑.

Now in this case, our two functions are 𝑦 and 𝑧. And they are in terms of π‘₯. So we can see that d𝑦 by d𝑧 must be equal to d𝑦 by dπ‘₯ divided by d𝑧 by dπ‘₯. So we’re going to need to begin by differentiating each of our functions with respect to π‘₯. The first derivative of seven π‘₯ is seven. And when we differentiate sin π‘₯, we get cos π‘₯. So we see that d𝑦 by dπ‘₯ here is seven plus four cos of π‘₯.

We also know that if we differentiate cos of π‘₯, we get negative sin of π‘₯. So that’s d𝑧 by dπ‘₯. It’s negative sin π‘₯. d𝑦 by d𝑧 is the quotient. It’s seven plus four cos of π‘₯ divided by negative sin π‘₯. But we’re not quite finished. We’re looking to find the derivative at the point where π‘₯ is equal to πœ‹ by six. So we’re going to substitute π‘₯ for πœ‹ by six in our expression. That’s seven plus four cos of πœ‹ by six over negative sin of πœ‹ by six.

Cos of πœ‹ by six is root three over two. And sin of πœ‹ by six is one-half. When we divide by one-half, that’s the same as multiplying the numerator by two. And so we see that the derivative of our function, seven π‘₯ plus four sin π‘₯, with respect to cos of π‘₯ plus one at π‘₯ equals πœ‹ by six is negative 14 minus four root three.

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