- #Quotient rule calculus low portable
- #Quotient rule calculus low iso
- #Quotient rule calculus low plus
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#Quotient rule calculus low iso
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#Quotient rule calculus low plus
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#Quotient rule calculus low portable
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The derivative of f is 2x times g of x, which This is going to be equal toį prime of x times g of x. And so now we're ready toĪpply the product rule. When we just talked about common derivatives. The derivative of g of x is just the derivative Just going to be equal to 2x by the power rule, and With- I don't know- let's say we're dealing with Now let's see if we can actuallyĪpply this to actually find the derivative of something. Times the derivative of the second function. In each term, we tookĭerivative of the first function times the second Plus the first function, not taking its derivative, Of the first one times the second function To the derivative of one of these functions, Of this function, that it's going to be equal Of two functions- so let's say it can be expressed asį of x times g of x- and we want to take the derivative If we have a function that can be expressed as a product Rule, which is one of the fundamental ways Personally, I don't think I would normally do that last stuff, but it is good to recognize that sometimes you will do all of your calculus correctly, but the choices on multiple-choice questions might have some extra algebraic manipulation done to what you found. If you are taking AP Calculus, you will sometimes see that answer factored a little more as follows: That gets multiplied by the first factor: 18(3x-5)^5(x^2+1)^3. Now, do that same type of process for the derivative of the second multiplied by the first factor.ĭ/dx = 6(3x-5)^5(3) = 18(3x-5)^5 (Remember that Chain Rule!) That gets multiplied by the second factor: 6x(x^2+1)^2(3x-5)^6 Your two factors are (x^2 + 1 )^3 and (3x - 5 )^6 Remember your product rule: derivative of the first factor times the second, plus derivative of the second factor times the first.
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