22–36. Derivatives Find the derivatives of th...

Question

22–36. Derivatives Find the derivatives of the following functions.

f(x) = x² cosh² 3x

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the derivative of H of X, which is equal to X 4th, multiplied by the hyperbolic cosine squared of 5 X with respect to X. So here we need to take a derivative of H of X with respect to X. So we're looking for H of X. And this is going to be equal to the derivative with respect to X. Of X to the 4th. Multiplied by the hyperbolic cosine squared of 5 X. So, in order to take this derivative, we'll need to apply our product rule. So, we call that for our product rule. H X is going to be equal to. We'll have the derivative of the first term, so we'll have the derivative with respect to X of X to the 4th. And then that gets multiplied by our second term in our product, which is going to be the hyperbolic cosine squared of 5 X. Then we'll have plus the first term in our product, which is gonna be X to the 4th, multiplied by the derivative with respect to X of the second term, which is going to be the hyperbolic cosine squared of 5 X. That we just need to take those derivatives. That means H X is going to be equal to. When I take the derivative with respect to X of X to the 4th, I get 4 X cubed, then that gets multiplied by the hyperbolic cosine squared. of 5 X. Then we'll have + X the 4th. And when we take, we multiply this by the derivative with respect to X of the hyperbolic cosine squared of 5 X, so we have to use our power rule to take that derivative, so we'll have 2, multiplied by the hyperbolic cosine. Of 5 X. And we'll actually need to apply our chain rule as well, so we need to multiply this by the derivative with respect to X. Of the hyperbolic cosine of 5 X. So, continuing to take derivatives, H X is going to be equal to. Or X cubed, multiplied by the hyperbolic cosine. Squad of 5 X. Plus, and here, we'll just rearrange, we'll bring the two out front, so we'll have 2 multiplied by X to the 4th, multiplied by the hyperbolic cosine of 5 X. Then this gets multiplied by the derivative spec X of the hyperbolic cosine of 5X. Recall that when you take a derivative of a hyperbolic cosine function, you get a hyperbolic sine function. So this is going to be multiplied by the hyperbolic sign of 5 X. But again, we'll need to apply our chain rule and multiply this by the derivative with respect to X of the argument of our hyperbolic function, which is going to be 5 X. So, taking our final derivative, we see that H of X is going to be equal to. 4 X cubed, multiplied by the hyperbolic cosine squared of 5 X. Plus 2 multiplied by X to the 4th, multiplied by the hyperbolic cosine of 5 X. Multiplied by the hyperbolic sign. Of 5 X. And then it gets multiplied by the derivative spec X of 5 X, which is just 5. And so finally simplifying our expression, we see that H of X is equal to. 4, multiplied by X cubed, multiplied by the hyperbolic cosine squared of 5 X. Plus, here we'll have 2 multiplied by 5, which will give me 10, that gets multiplied by X to the 4th, multiplied by the hyperbolic cosine of 5 X, multiplied by the hyperbolic sine. Of 5 acts. So this here is the answer that we're looking for, for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

22–36. Derivatives Find the derivatives of the following functions.

f(x) = x² cosh² 3x

Key Concepts

Derivatives

Product Rule

Hyperbolic Functions

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