Problem

Differentiate. $ F(t) = \frac {At}{Bt^2 + Ct^3…

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Problem 23 Easy Difficulty

Differentiate.

$ f(x) = \frac {x^2e^x}{x^2 + e^x} $

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Video Transcript

All right here we have a fun problem to solve our function in fred here at the bottom is x, squared to the x over x, squared plus e, to the x and notice that there is a product in the numerator in the whole function itself is a Quotient so notice we're going to need to use both the product rule and the potient rule just reminder of the product rule. If i take the derivative of f times g, then it's the derivative of f times g plus g prime times f. If i do quotient rule instead, it's a derivative of the top times the bottom nicest derivative of the bottom times, the top all over the bottom square. It so we're going to use both rules to solve. So here we go. Okay, so f, prime of x, is going to be basically a quotient rule and overall, that's my! U and that's my v so we're going to first take the derivative of the top well, the top is itself, and it's going to be a big long fraction on the top is itself a product, so we're going to first start off by doing u, prime and That'S going to be product role, so derivative of the first function times the second point, so we get 2 x times e to the x plus a derivative of the second, which is just itself times the first. So that's prime! Then we multiply by b and that's just the denominator, so we can write it as it. Then we have to subtract and we're going to take the derivative of the denominator. Derivative of x, squared is 2 x by power rule and then a derivative of e to the x is itself. Then we multiply by the numerator and that's our basically. We did this part so far, prime v minus v, prime. U we still need to divide by the denominator, squared so we're going to get over x, squared plus e to the x quantity squared all right, so this is massawe have completed it, but let's have fun and see if we can clean it up a bit. So let's go ahead and distribute the top and see what we get okay, so at prime of x, i'm going to foil first, the top, so i'm gonna do multiply the first. So that gives me 2 x cube e to the x. Then i go at the 2 outer ones so, plus 2 x e to the 2 x this x to the x times of the x i e to the 2 x. Then i do the inner and that's these 2. So i'm going to get plus e to the x x to the fourth and that's a plus and then finally, the last so i'll get plus x, squared e to the 2, all right. Okay, now we're going to go ahead, so that was all just spoiling. These first 2 terms, now let's go ahead and distribute over this expert to the x over both parts and not forget about the minus sign. So i gives us minus 2 x, cube e to the x right, and then we have a minus x, squared e to the 2 x because e to the x times e to the x, i e to the 2 x and then our bottom will just rewrite. As is, and let's see if we are fortunate enough to have anything, cancel out, because that would be amazing: oh yea, look 2 x, cube to the x, cancels the minus 2 x cube to the x, and i and other yes expert the 2 x cancels expert. I the 2 x. This is really good news, so we're almost there. Okay, so f, prime of x. Let'S just see if there's anything common in those 2 remaining terms, it looks like they both have at least 1 x and an et the x, so that will leave behind 2 e to the x plus x, cubed, double check. It multiplies out right and that looks good all over x, squared plus e to the x quantity squared, and that should do us. I think, that's as clean as we could get for this derivative. Okay, hopefully that helped to have an a.

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Problem

Differentiate. $ F(t) = \frac {At}{Bt^2 + Ct^3…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Joseph Lentino

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

Joseph Lentino

James Stewart

Calculus: Early Transcendentals