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

# Question Video: Differentiating Trigonometric Functions Using the Product Rule Mathematics • Third Year of Secondary School

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Find the first derivative of π¦ : π¦ = 16 csc 3π₯ sec 4π₯.

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### Video Transcript

Find the first derivative of π¦ where π¦ is equal to 16 csc of three π₯ multiplied by the sec of four π₯.

The question wants us to find the first derivative of π¦, and it gives us an expression for π¦. We can see that π¦ is a function in π₯. So when the question asks for the first derivative of π¦, it means the first derivative of π¦ with respect to π₯. And thereβs actually a lot of different ways of evaluating this expression. For example, we could use our trigonometric identities to rewrite the csc of three π₯ and the sec of four π₯ in our denominator as the sin of three π₯ and the cos of four π₯, respectively. We could then differentiate this by using the product rule and the chain rule or the product rule and the general power rule or the product rule and the quotient rule. All of these methods would work. However, thereβs actually an easier method. We need to notice that we know how to different the csc of three π₯ and the sec of four π₯.

This means that π¦ is written as the product of functions we know how to differentiate. So we can directly differentiate this by using the product rule. So letβs start by recalling the product rule. If π¦ is the product of two differentiable functions, π of π₯ multiplied by π of π₯, then π¦ prime, the derivative of π¦ with respect to π₯, will be equal to π prime of π₯ times π of π₯ plus π prime of π₯ times π of π₯. And we can see that our expression for π¦ is in exactly this form. We set π of π₯ to be 16 times the csc of three π₯ and π of π₯ to be the sec of four π₯.

Now, to use the product rule, we just need to find expressions for π prime of π₯ and π prime of π₯. Letβs start with π prime of π₯. Thatβs the derivative of 16 csc of three π₯ with respect to π₯. The easiest way to evaluate this derivative is to recall the following result for differentiating reciprocal trigonometric functions. For any real constant π, the derivative of the csc of ππ₯ with respect to π₯ is equal to negative π times the csc of ππ₯ multiplied by the cot of ππ₯. In this case, we can see the value of our constant π will be three. So by setting our value of π equal to three and remembering we need to multiply by 16, we get π prime of π₯ is equal to 16 multiplied by negative three csc of three π₯ multiplied by the cot of three π₯.

And of course, we can simplify this. 16 multiplied by negative three is equal to negative 48. So we were able to find an expression for π prime of π₯. Letβs now move on to finding an expression for π prime of π₯. Thatβs the derivative of the sec of four π₯ with respect to π₯. And we can evaluate this derivative by recalling the following trigonometric derivative result. For any real constant π, the derivative of the sec of ππ₯ with respect to π₯ is equal to π times the sec of ππ₯ multiplied by the tan of ππ₯. This time, we can see our value of the constant π is equal to four. So by setting the value of π equal to four, we get π prime of π₯ is equal to four times the sec of four π₯ multiplied by the tan of four π₯.

Now that we found expressions for π prime of π₯ and π prime of π₯ and we know π of π₯ and π of π₯, weβre ready to find an expression for π¦ prime. So weβll clear some space and substitute these expressions into our product rule. This gives us the first derivative of π¦ with respect to π₯ is equal to negative 48 csc of three π₯ times the cot of three π₯ multiplied by the sec of four π₯ plus four sec of four π₯ times the tan of four π₯ multiplied by 16 times the csc of three π₯. And we want to simplify this expression. First, weβll take out the shared factor of 16 from both terms and the shared factor of the csc of three π₯ from both terms and also the shared factor of the sec of four π₯ from both terms.

So weβre taking out our factor of 16 times the csc of three π₯ multiplied by the sec of four π₯. Remember that weβve only taken out a factor of 16 from our original coefficient, and we know negative 48 divided by 16 is negative three. So this gives us negative three times the cot of three π₯. In our second term, weβre just left with four times the tan of four π₯. And this just leaves us with the following expression, and we could just leave our answer like this. However, thereβs one more piece of simplification weβre going to do. Inside of our parentheses, weβre going to rearrange our two terms. And this gives us our final answer.

Therefore, we were able to show if π¦ is equal to 16 csc of three π₯ times the sec of four π₯, then the first derivative of π¦ with respect to π₯ is equal to 16 csc of three π₯ times the sec of four π₯ multiplied by four tan of four π₯ minus three cot of three π₯.

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