Video: Finding the Derivative of a Trigonometric Function Using the Chain Rule

Find the derivative of the function 𝐽(πœƒ) = tanΒ² π‘›πœƒ.

02:43

Video Transcript

Find the derivative of the function 𝐽 of πœƒ is equal to the tan squared of 𝑛 times πœƒ.

The question gives us a function 𝐽 in terms of πœƒ. And it wants us to find the derivative of this function. Since our function is given in terms of πœƒ, we want to find the derivative of 𝐽 with respect to πœƒ. So we want to find the derivative with respect to πœƒ of the tangent squared of π‘›πœƒ. There’s a few different ways of doing this. We’re going to do this by realizing that the tan squared of π‘›πœƒ is just equal to the tan of π‘›πœƒ multiplied by the tan of π‘›πœƒ. Now, we can see the derivative that we need to evaluate is the derivative of the product of two functions. So we can do this by using the product rule. We recall the product rule tells us, for two functions 𝑒 and 𝑣, the derivative of 𝑒 times 𝑣 with respect to πœƒ is equal to 𝑣 times d𝑒 by dπœƒ plus 𝑒 times d𝑣 by dπœƒ.

So to apply this rule, we’ll set both 𝑒 and 𝑣 equal to the tan of 𝑛 times πœƒ. So to use the product rule, we’re going to need to find expressions for d𝑒 by dπœƒ and d𝑣 by dπœƒ. Both of these are equal to the derivative of the tan of π‘›πœƒ with respect to πœƒ. And we know, for any constant π‘Ž, the derivative of the tan of π‘Ž times πœƒ with respect to πœƒ is equal to π‘Ž times the sec squared of π‘Ž times πœƒ. And we can then use this to evaluate both of our derivatives. We get that d𝑒 by dπœƒ and d𝑣 by dπœƒ are both equal to 𝑛 times the sec squared of 𝑛 times πœƒ.

We’re now ready to evaluate the derivative of the tan squared of 𝑛 times πœƒ with respect to πœƒ by using the product rule. We get that it’s equal to 𝑣 times d𝑒 by dπœƒ plus 𝑒 times d𝑣 by dπœƒ. Substituting our expressions for 𝑒, 𝑣, d𝑒 by dπœƒ, and d𝑣 by dπœƒ, we get the d𝐽 by dπœƒ is equal to the tan of 𝑛 times πœƒ multiplied by 𝑛 times the sec squared of π‘›πœƒ plus the tan of π‘›πœƒ times 𝑛 times the sec squared of π‘›πœƒ. And we can see that both of the expressions here which we’re adding together are equal.

So to add these two terms together, we can just multiply one of the terms by two. And we can rearrange this to give two 𝑛 times the sec squared of π‘›πœƒ multiplied by the tan of π‘›πœƒ. And remember, this was an expression for the first derivative of our function 𝐽 of πœƒ. So this is equal to 𝐽 prime of πœƒ. Therefore, we’ve shown that 𝐽 of πœƒ is equal to the tan squared of π‘›πœƒ. Then 𝐽 prime of πœƒ is equal to two 𝑛 times the sec squared of π‘›πœƒ times the tan of π‘›πœƒ.

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