mustry
any one help me abt calculus



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Nimi404
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Wat do mean

Nimi404
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Calculate wat

mustry
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If we have a product like
y = (2x2 + 6x)(2x3 + 5x2)

AravindG
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you need differential?

mustry
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d(uv)dx=udvdx+vdudx

AravindG
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if you need to differentiate use product rule

Nimi404
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Hey Aravin you again

AravindG
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yep apply that

AravindG
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@Nimi404 calculus is higher mathematics it doesnt mean calculate :P

mustry
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how do derive this formula

AravindG
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you mean you need to derive product rule ?

Nimi404
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Ooh then

mustry
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Find the derivative of
y = (x3 − 6x)(2 − 4x3)

AravindG
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confused ...can u be clear ?

mustry
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please help me this

mustry
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Find the derivative of
y = (x3 − 6x)(2 − 4x3)

Yahoo!
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u know Product Rule ? @mustry

mustry
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yap

AravindG
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dw:1356793984995:dw

AravindG
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where i took x^36x as first function and 24x^3 as second function

mustry
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then

AravindG
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then used product rule

mustry
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thankx but can u derive it product rule?

Callisto
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Derive the product rule?

mustry
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yes

Callisto
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Okay...*Take a deep breath*\[\frac{d}{dx}(uv)\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)\color{red}{+u(x+h)v(x)u(x+h)v(x)}u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]v(x)[u(x+h)u(x)]}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]}{h}\lim_{h \rightarrow 0}\frac{v(x)[u(x+h)u(x)]}{h}\]\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)+v(x)}{h}v(x)\lim_{h \rightarrow 0}\frac{u(x+h)u(x)}{h}\]\[=u(x)v'(x)+u'(x)v(x)\]