In Exercises 41–58, find dy/dt.

y = tan²(sin³...

Question

In Exercises 41–58, find dy/dt.

y = tan²(sin³(t))

Accepted Answer

Welcome back everyone. In this problem, we want to find the derivative of Y equals cut cubed of 4th of X. Now, before we start to solve for the derivative, let's make a note of a few things here, OK? Now, in this problem, and this equation, Y equals of 4 X. What we're really saying here if we were to rewrite this expression is that we're cubing the cotangent of this expression inside. So in other words, we're finding the cotangent of the cosine of X are raised to the 4th, OK. And then we're cuing that result. OK? This is one way that we can think about it. And now when we look at it this way, it might seem a little more obvious to tell that we will need to use the chain rule of differentiation because we're dealing with a composite function and we'll also need to use the power rule. Now what do we know about both of those? Well, recall, OK, that the chain rule, or first the poor rule of differentiation says that if we're differentiating a term. Raised to a power, OK, X to the N, then its derivative will be equal to. N X N minus 1. And we also know that if we're differentiating a composite function, OK, let's say that it's F of G of X. OK, then it's derivative will be equal to the derivative of F of geoffX, OK, multiplied by the derivative of G of X. In other words, we bring down the power or we differentiate the outer function and then multiply it by the derivative of the inner function, OK? So now, in this case, because we're also dealing with trigonometric functions here. We can also make note that the derivative of the cotangent, OK, the derivative of the cotangent of X. Will be equal to negative core of X, OK? And we can also make note that the derivative of a cosine of X. Will be equal to the or the negative value of the sign of X. So now that we have a few of these in our wool box as it were, it might help us to differentiate uh why a little bit easier. Because now, OK, if we apply our rules here, since we know that Y is a composite function differentiating it with the chain rule, then the derivative of Y, OK, is going to be equal to first we're gonna bring down the power 3 multiplied by a cotangent to the 3 minus 1. In other words, we subtract 1 from the power of the cosine of X sorry, the cosine. Of x to the 4th multiplied by the derivative of that inner function. In other words, the derivative of the cotangent of co4th of X. So now, in order for us to find the derivative of Y, we need to then differentiate this expression. So what is that going to be? Well, OK, if we differentiate here, so we find a derivative of the cotangent ofga4Xs, OK. Then again, we can apply the chain rule, which means we're going to differentiate the cotangent as negative core C can squared of across for. Of X, OK, multiplied by the derivative of the inner function. So that's the derivative of cost 4th of X. We have a bit more to do because we know we need to differentiate the cost of X, OK. So now we're right back, negative coc can squad of 4 of X, OK. No, we're gonna bring down the power. So that's 4 multiplied by the cosine of 4 minus 1 of X multiplied by the derivative of the cosine of X. OK. Again, we're differentiating our inner function. So in a sense, we've expressed CA4 to the X as a composite function. C C X R to the 4th, and now that's what we're we we've differentiated here. OK, in the previous line. No, by differentiating the cosine of X, then if we go ahead, we're gonna have right back what we had before. OK, cause cube of X 4 minus 1 is 3. I know the derivative of the cosine of X, like we said earlier is negative sin X. So now if we simplify here, OK, then we should get this. OK. To be equal to, well, here, notice that before in our line, OK, we have negative multiplied by negative, so that's positive, OK. So this is really gonna be 4 multiplied by coi can squared of the cosine to the 4th of X, OK? Multiplied by cost cubs of X multiplied by sin X, OK? So this is the derivative we wanted. Now that we have that we can get back to our derivative of Y, which means then. Alright I think about the terms that we had before, so it's 3 multiplied by the, the cotangent squared of cost 4th of X. Multiplied by our derivative that we just found. And for simplicity, let me just copy and paste it here. OK, in the interest of time. But now, this is the the derivative that we're finding. All we need to do now is to simplify, OK. So, now, when we do that, 3 multiplied by 4 is 12, OK. We write cost cubed X here, OK? Taking that term, multiplied by squared of for X, for of X, OK. Multiplied by cosquared of cost of X, OK. Multiplied by the sine of X. And this will be the derivative of our function Y. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 41–58, find dy/dt.

y = tan²(sin³(t))

Key Concepts

Chain Rule

Derivative of Trigonometric Functions

Power Rule

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