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any one help me abt calculus

Mathematics
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Wat do mean
Calculate wat
If we have a product like y = (2x2 + 6x)(2x3 + 5x2)

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Other answers:

you need differential?
d(uv)dx=udvdx+vdudx
if you need to differentiate use product rule
Hey Aravin you again
yep apply that
@Nimi404 calculus is higher mathematics it doesnt mean calculate :P
how do derive this formula
you mean you need to derive product rule ?
Ooh then
Find the derivative of y = (x3 − 6x)(2 − 4x3)
confused ...can u be clear ?
please help me this
Find the derivative of y = (x3 − 6x)(2 − 4x3)
u know Product Rule ? @mustry
yap
|dw:1356793984995:dw|
where i took x^3-6x as first function and 2-4x^3 as second function
then
then used product rule
thankx but can u derive it product rule?
Derive the product rule?
yes
Okay...*Take a deep breath*\[\frac{d}{dx}(uv)\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)-u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)v(x+h)\color{red}{+u(x+h)v(x)-u(x+h)v(x)}-u(x)v(x)}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]-v(x)[u(x+h)-u(x)]}{h}\]\[=\lim_{h \rightarrow 0}\frac{u(x+h)[v(x+h)+v(x)]}{h}-\lim_{h \rightarrow 0}\frac{v(x)[u(x+h)-u(x)]}{h}\]\[=\lim_{h \rightarrow 0}u(x+h)\lim_{h \rightarrow 0}\frac{v(x+h)+v(x)}{h}-v(x)\lim_{h \rightarrow 0}\frac{u(x+h)-u(x)}{h}\]\[=u(x)v'(x)+u'(x)v(x)\]

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