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mustry
 2 years ago
Best ResponseYou've already chosen the best response.0If we have a product like y = (2x2 + 6x)(2x3 + 5x2)

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2you need differential?

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2if you need to differentiate use product rule

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2@Nimi404 calculus is higher mathematics it doesnt mean calculate :P

mustry
 2 years ago
Best ResponseYou've already chosen the best response.0how do derive this formula

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2you mean you need to derive product rule ?

mustry
 2 years ago
Best ResponseYou've already chosen the best response.0Find the derivative of y = (x3 − 6x)(2 − 4x3)

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2confused ...can u be clear ?

mustry
 2 years ago
Best ResponseYou've already chosen the best response.0Find the derivative of y = (x3 − 6x)(2 − 4x3)

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0u know Product Rule ? @mustry

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2where i took x^36x as first function and 24x^3 as second function

AravindG
 2 years ago
Best ResponseYou've already chosen the best response.2then used product rule

mustry
 2 years ago
Best ResponseYou've already chosen the best response.0thankx but can u derive it product rule?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0Derive the product rule?

Callisto
 2 years ago
Best ResponseYou've already chosen the best response.0Okay...*Take a deep breath*\[\frac{d}{dx}(uv)\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)\color{red}{+u(x+h)v(x)u(x+h)v(x)}u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]v(x)[u(x+h)u(x)]}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]}{h}\lim_{h \rightarrow 0}\frac{v(x)[u(x+h)u(x)]}{h}\]\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)+v(x)}{h}v(x)\lim_{h \rightarrow 0}\frac{u(x+h)u(x)}{h}\]\[=u(x)v'(x)+u'(x)v(x)\]
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