Find y'' for the following functions.
y = x sin x

Question

Find y'' for the following functions.

y = x sin x

Accepted Answer

Welcome back, everyone. In this problem, we want to find the 2nd derivative Y for the function Y equals X 5 multiplied by the cosine of X. Now to find the second derivative of Y, what it basically means is that we just need to differentiate Y twice. So let's first find Y the first derivative and then differentiate that again to get Y, the second derivative. Now, given Y, which is equal to X 5th multiplied by the cosine of X, we can tell that Y is our product. What do we know about differentiating products? Well, recall the product rule of differentiation that tells us if Y is equal to a product u multiplied by V where U and V are differentiable functions. Then the derivative of Y is going to be equal to UV plus VU. In other words, we're going to differentiate both functions and multiply them by the opposite original function. So in this case, If we can differentiate U and V, differentiate both our terms, we can plug them into the product tool to to find the first derivative, OK? Now here if we let you. OK, if we let U be equal to x 5th, then the derivative of U equals 5 X4 using the power rule of differentiation. Likewise, OK. If we let V be equal to the cosine of X, then V the derivative is going to be negative sine X because that's the derivative of class X. Now that we have U and V, we can plug them into the product tool which then tells us that Y will be equal to U multiplied by V. that's 5 X to the 4th multiplied by the cosine of X plus. OK. You multiplied by VX to the 5th multiplied by negatives X. And now when we simplify that, OK. Here, X 5 multiplied by negative. Xin X is gonna be equal to negative X 5th sin X. So I can rewrite that here, OK? So let me just fix that up. That as minus x to the 5th multiplied by the sine of X. So we have the first derivative of Y. To find the second derivative, we're going to differentiate Y. Now, notice here, Y has two terms and both of them are products. So we're going to apply the product rule for both of our terms, OK? So now, let me put this in blue here. So we're solving for the second derivative. OK. Now, just to make it easier for us to follow, let's set. F of X, OK, to be equal to 5 x 4 multiplied by the cosine of X. And let's say GX, OK, to be equal to. Let's say G of X to be equal to negative x 5 multiplied by the sine of X. No, if we're differentiating both of these again, we'd use the power rule, OK? So we know the power rule here. If we let, let's call U be equal to 5 X to the 4th, then the derivative is going to be equal to 20 X cubed, OK? And if we say V is the coresine of X, then the derivative of V will be negative sin X, OK? So, here we can use this for our first function. Because no, that means then that the derivative of the first function. Is going to be equal. So we're gonna multiply these. So UV it's gonna be -5 X to the fourth multiplied by sin x plus VU. So that's 20 X cubed multiplied by the cosine of X. So we have the derivative of our first term, OK, in our first derivative. Let's find the derivative of our second term in the first derivative. So for it to be negative X to the 5th multiplied by a sin X, we can say U is negative X to the 5th, which means U is going to be -5 X to the 4th. And, we can let V be equal to sin X. So V is gonna be equal to the cosine of X. That's the derivative of the sin of X, OK? So, no, that means then that the derivative of G of X. Is going to be equal to -5x the 4th multiplied by the sign of x. Plus negative x to the 5th multiplied by the cosine of X. So have you seen what we've done here? Essentially all we did was that we applied the same product rule we used to find the first derivative to differentiate each of the terms in the second derivative because both of them are products. Know that we've been able to do that, OK? Now that we found those derivatives, then all we need to do now is to combine our results. Because since we've differentiated them independently, separately, then it's going Y double prime is going to be equal to 20 x cubed plus x. Minus 5x the 4th sine x the derivative of the 1st expression. Plus 5 x 4 x minus X 5th plus X, the derivative of the second expression. And know if we try to simplify here, OK. Notice we have 20, -5 X to the 4th, sine x minus 5 X to the 4th. Sin X, OK. So, no, that means we're gonna have negative X to the 5th plus X. minus 10 X 4th, sin X, OK? plus 20 X cubed plus X. This is the result after we combine all our terms. Thus, this is the second derivative of y. Thanks a lot for watching everyone. I hope this video helped.

Find y'' for the following functions.y = x sin x

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Find y'' for the following functions.
y = x sin x