## mustry 2 years ago any one help me abt calculus

1. Nimi404

Wat do mean

2. Nimi404

Calculate wat

3. mustry

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

4. AravindG

you need differential?

5. mustry

d(uv)dx=udvdx+vdudx

6. AravindG

if you need to differentiate use product rule

7. Nimi404

Hey Aravin you again

8. AravindG

yep apply that

9. AravindG

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

10. mustry

how do derive this formula

11. AravindG

you mean you need to derive product rule ?

12. Nimi404

Ooh then

13. mustry

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

14. AravindG

confused ...can u be clear ?

15. mustry

16. mustry

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

17. Yahoo!

u know Product Rule ? @mustry

18. mustry

yap

19. AravindG

|dw:1356793984995:dw|

20. AravindG

where i took x^3-6x as first function and 2-4x^3 as second function

21. mustry

then

22. AravindG

then used product rule

23. mustry

thankx but can u derive it product rule?

24. Callisto

Derive the product rule?

25. mustry

yes

26. Callisto

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)$