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# If $y = f(u)$ and $u = g(x),$ where $f$ and $g$ possess third derivatives, find a formula for $d^3 y/dx^3$ similar to the one given in Exercise 99.

## $\frac{d^{3} y}{d x^{3}}=\frac{d^{3} y}{d u^{3}}\left(\frac{d u}{d x}\right)^{3}+3 \frac{d^{2} y}{d u^{2}} \frac{d u}{d x} \frac{d^{2} u}{d x^{2}}+\frac{d y}{d u} \frac{d^{3} u}{d x^{3}}$

Derivatives

Differentiation

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we have this equation for a second derivative from problem number 99. And now we're going to find its derivative to give us an equation for the third derivative. And the way we're going to find its derivative is will use the product rule on this and will use the product rule on this. So let's start with the 1st 1 So we have the first D squared. Why do you squared times the derivative of second? Now, when we take the derivative of this, we have to use the chain rule. So we bring down the to and then we raised the utx to the first and then we multiply by the derivative of Dut X, which would be d squared You d x squared the second derivative. So so far we have in our product rule the first times a derivative of the second. And let's do plus the second dese do you d X squared times the derivative of the first and to take the derivative of D squared. Why do you squared? We're going to have 1/3 derivative t cubed Why d you cubed times a derivative of the inside which would be dut X Okay, so that took us through the first product rule from the first product. Now we move on to the second product rule. So plus, now we're looking at the second product and we have the first do I. D. You times a derivative of the second. So we're taking the derivative of a second derivative That would be 1/3 derivative. So d cubed you d x cubed plus the second D squared you d X squared times the derivative of the first. So we're taking the derivative of Do I Do you? So that's going to be second derivative of why times the derivative of the inside, which is dut X. Now we want to see if there's anything that we can combine or simplify. So noticed that our notice that our first term and our last term both contain the same variable expressions. The second derivative of why, with respect to you the second derivative of you with respect to X and a d u d X and so we have two of them in the first term. We have one of them in the second or the last term, and so we can add those together and we have three of them. So we have three times t squared. Why do you squared times? Do you? D x times d squared you d X squared. Now, if we look closely at our middle two terms, we noticed that they are not the same. And one thing we could do is we have do you d x squared times to utx So we can write that as dut x cubed So we have dut X cubed times the third derivative of why with respect to you And then finally this one term that we were not able to combine with anything Do you? Why do you times that they're derivative of you Whips third derivative of you with respect to X not a formula. I wanted memories I don't know about you

Oregon State University

Derivatives

Differentiation

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