Question Video: Differentiating Trigonometric Functions Using the Chain Rule | Nagwa Question Video: Differentiating Trigonometric Functions Using the Chain Rule | Nagwa

Question Video: Differentiating Trigonometric Functions Using the Chain Rule Mathematics

Find d𝑦/dπ‘₯, given that 𝑦 = (βˆ’3 cot 5π‘₯ + 7)⁴.

03:08

Video Transcript

Find d𝑦 by dπ‘₯, given that 𝑦 is equal to negative three times the cot of five π‘₯ plus seven all raised to the fourth power.

We’re given 𝑦 as a function in π‘₯ and asked to find d𝑦 by dπ‘₯. That’s the first derivative of 𝑦 with respect to π‘₯. To do this, we’re going to first need to take a look at our function for 𝑦. We can see that 𝑦 is a trigonometric function all raised to the fourth power. In other words, 𝑦 is the composition of two functions. And we know we can differentiate the composition of two functions by using the chain rule. However, because our outer function is a power function, we can also do this by using the general power rule. It doesn’t matter which method we use; both will give us the same answer. It’s all personal preference. We’re going to use the general power rule.

So let’s start by recalling the general power rule. The general power rule tells us for any real constant 𝑛 and differentiable function 𝑓 of π‘₯, if we have 𝑦 is equal to 𝑓 of π‘₯ all raised to the 𝑛th power, then d𝑦 by dπ‘₯ will be equal to 𝑛 times 𝑓 prime of π‘₯ multiplied by 𝑓 of π‘₯ all raised to the power of 𝑛 minus one. And we can see that 𝑦 is exactly written in this form, with our function 𝑓 of π‘₯ as our inner function, negative three cot of five π‘₯ plus seven, and our exponent 𝑛 equal to four. So, by setting 𝑓 of π‘₯ to be our inner function and 𝑛 equal to four, we can try and find d𝑦 by dπ‘₯ by using the general power rule.

To do this, we’re going to need to find an expression for 𝑓 prime of π‘₯. That’s the derivative of negative three cot of five π‘₯ plus seven with respect to π‘₯. And we can differentiate this directly if we recall one of our rules for differentiating the reciprocal trigonometric functions. We know for any real constant π‘Ž, the derivative of the cot of π‘Žπ‘₯ with respect to π‘₯ is equal to negative π‘Ž times the csc squared of π‘Žπ‘₯. We can use this to differentiate the first term in 𝑓 of π‘₯. We just need to set our value of π‘Ž equal to five.

So by setting our value of π‘Ž equal to five and remembering we need to multiply this expression by negative three, differentiating our first term, we get negative three times negative five csc squared of five π‘₯. And of course, we know that seven is a constant, so its rate of change with respect to π‘₯ will be equal to zero. So 𝑓 prime of π‘₯ is just negative three multiplied by negative five csc squared of five π‘₯. Now, all we need to do is simplify this expression. Negative three times negative five is 15, so 𝑓 prime of π‘₯ is 15 csc squared of five π‘₯. We now have a value of 𝑛, an expression for 𝑓 prime of π‘₯, and 𝑓 of π‘₯, so we can use the general power rule to find d𝑦 by dπ‘₯.

Substituting 𝑛 is equal to four and our expressions for 𝑓 prime of π‘₯ and 𝑓 of π‘₯ into our general power rule, we get d𝑦 by dπ‘₯ is equal to four times 15 csc squared of five π‘₯ multiplied by negative three cot of five π‘₯ plus seven all raised to the power of four minus one. And we can simplify this. First, in our exponent, four minus one is equal to three. We can also simplify our coefficient. Four multiplied by 15 is equal to 60. So by using this and rearranging our factors, we get the following expression as our final answer.

Therefore, by using the general power rule, we were able to show if 𝑦 is equal to negative three cot of five π‘₯ plus seven all raised to the fourth power, then d𝑦 by dπ‘₯ will be equal to 60 times negative three cot of five π‘₯ plus seven all cubed multiplied by the csc squared of five π‘₯.

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