Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Quotient Rule | Nagwa Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Quotient Rule | Nagwa

# Question Video: Differentiating Functions Involving Trigonometric Ratios Using the Quotient Rule Mathematics • Second Year of Secondary School

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If π¦ = (7 sin 2π₯)/(2 β 2 cos 2π₯), find dπ¦/dπ₯.

03:40

### Video Transcript

If π¦ equals seven sin two π₯ divided by two minus two cos two π₯, find dπ¦ by dπ₯.

Weβre looking to find dπ¦ by dπ₯, the derivative of a function which is itself the quotient of two differentiable functions. This means we can use the quotient rule to find this derivative. The quotient rule says that the derivative of the quotient of two differentiable functions π’ and π£ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ over π£ squared. So letβs define π’ and π£ in our question.

The numerator of our fraction is seven sin two π₯. So weβll let π’ be equal to this. Then, π£ is equal to two minus two cos of two π₯. Our next job is to evaluate dπ’ by dπ₯ and dπ£ by dπ₯. So next, we state the general results for the derivative of the sine and cosine function. The derivative of sin ππ₯ is π cos π of π₯. And the derivative of cos ππ₯ is equal to negative π sin of ππ₯. We can say then that dπ’ by dπ₯ must be equal to seven times two cos of two π₯, which is just 14 cos of two π₯. And since the derivative of a constant is zero, dπ£ by dπ₯ is simply negative two times negative two sin of two π₯, which is four sin of two π₯.

Our next job is to substitute everything into the formula for the quotient rule. We have π£ times dπ’ by dπ₯. Thatβs two minus two cos of two π₯ times 14 cos of two π₯ minus π’ times dπ£ by dπ₯. Thatβs seven sin of two π₯ times four sin of two π₯, which is, of course, all over π£ squared, two minus two cos of two π₯ squared. Weβll look to distribute the parentheses on the numerator of our fraction. When we do, we get 28 cos of two π₯ minus 28 cos squared of two π₯ minus 28 sin squared of two π₯. And for now, weβll keep the numerator as two minus two cos of two π₯ all squared. We can actually divide through by a factor of four. And when we do, we obtain dπ¦ by dπ₯ to be equal to seven cos of two π₯ minus seven cos squared two π₯ minus seven sin squared two π₯ all over one minus cos of two π₯ all squared. This still looks rather nasty though. So letβs look to use some trigonometric identities to simplify it further. Letβs clear some space.

The first thing weβre going to do is factor negative seven on the numerator. And now, you might be able to see why this is a useful step to take. We now have an identity that relates cos squared π₯ and sin squared π₯. That is cos squared π₯ plus sin squared π₯ is equal to one. So we see that dπ¦ by dπ₯ is equal to negative seven times negative cos of two π₯ plus one over one minus cos of two π₯ squared. Letβs rewrite the numerator as one minus cos of two π₯. And then, we see that we can simplify by dividing both the numerator and the denominator by one minus cos of two π₯.

Next, we quote the double angle identity, βcos of two π₯ is equal to one minus two sin squared π₯.β If we rearrange this, we see that two sin squared π₯ is equal to one minus cos of two π₯. So dπ¦ by dπ₯ is now negative seven times one over two lots of sin squared π₯. Weβll manipulate this a little and write it as negative seven over two times one over sin squared π₯ because then we can use the identity one over sin π₯ is equal to csc of π₯. So one over sin squared π₯ is equal to csc squared π₯.

And we, therefore, see that dπ¦ by dπ₯ is equal to negative seven over two csc squared π₯.

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