# Reverse Chain Rule Help

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Hi I'm wondering if anybody can explain the reverse chain rule to me? I don't mean the whole U-Substitution thing, I mean the "shortcut" way.  The formula is f'(g(x)) · g'(x) dx = f(g(x)) + c.

Also, we haven't learned calculus and trig or calculus and logs yet so please just use basic equations in any examples. Thank  you!!!!

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This so-called reverse chain rule is u-substitution. it's just written differently. Also I'm pretty sure you have seen results based on this rule, even if it weren't introduced as such. Quick remark about notation, f'(x) is the derivative of f(x) and so f'(g(x)) is (f' º g) (x).

Say f(x) = x³, f'(x) = 3x², g(x) = 5x - 7 and g'(x) = 5. From here, we see f'(g(x)) g'(x) = 3(5x - 7)² × 5. So ∫﻿15(5x - 7)² dx = (5x - 7)³ + c. Applying the reverse chain rule might become so automatic in integration of polynomials that sometimes you forgot you are applying a rule. Maybe you were taught that if a factor is inside the base, divide by the factor when taking the integral. But this comes from this very rule of reverse chain rule/u substitution.

Take another example, ∫﻿e^(2x + 1) dx. f(x) = f'(x) = e^x. g(x) = 2x + 1; g'(x) = 2. Then we know that ∫﻿2 * e^(2x + 1) dx = e^(2x + 1) + c, so ∫﻿e^(2x + 1) dx = 1/2 * e^(2x + 1) + c.

Final example,  ∫﻿x(x² + 3)^4 dx = (1/10) * (x² + 3)^5 + c because  ∫5﻿(x² + 3)^4  * (2x) dx = (x² + 3)^5 + c. Where f(x) = x^5, f'(x) = 5x^4, g(x) = x² + 3, and g'(x) = 2x.

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Omg thank you so much for always helping me hahaha. I was trying to learn that by myself and totally got confused. I understand how u-substitution works now so thank you so much ## Join the conversation

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