Question

guys please help me i'am really having a hard time in differential calculus

Answer 1

derivatives?

Answer 2

can you be more accurate about your problem?

Answer 3

x^2sinx+2xcosx-2sinx

Answer 4

uhm differentiation of inverse trigonometric functions

Answer 5

r u trying to differentiate: x^2sinx+2xcosx-2sinx ?

Answer 6

yes differentiate thanks

Answer 7

sin(u)= cos(u)

Answer 8

cosu=sinu

Answer 9

@renzarabit I can help you some. First off though:
sin(u)=cos(u) is a false statement. A true statement would be:
\[\frac{d}{du}\sin(u)=\cos(u)\]
Which means: "The derivative of sin(u) with respect to u equals cos(u)."
This is a true statement.
Also:
\[\frac{d}{du}\cos(u)=-\sin(u)\]
"The derivative of cos(u) with respect to u is Negative sin(u)."
These are two simple trig derivatives you should devote to memory.

Answer 10

oh thanks :)

Answer 11

Back to your original problem:
\[\frac{d}{dx} \left[ x^2 \sin(x)+ 2x \cos(x)-2 \sin(x)\right]\]
You need the product rule. The product rule says that if you have two functions, f(x) and g(x), multiplied together and you take the derivative you get:
\[\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)\]
Where f'(x) means d/dx(f(x)).
So apply this to the first term you have:
\[\frac{d}{dx}x^2 \sin(x) = (2x)\sin(x)+x^2(\cos(x))\]
Does that make sense?

Answer 12

two teachers at the same time for this is too many. I leave it up to you @malevolence19
Goos luck @renzarabit !

Answer 13

*I meant good luck

Answer 14

Thanks @CarlosGP :P

Answer 15

thanks @CarlosGP :)))