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Mathematics 13 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

derivatives?

OpenStudy (anonymous):

can you be more accurate about your problem?

OpenStudy (anonymous):

x^2sinx+2xcosx-2sinx

OpenStudy (anonymous):

uhm differentiation of inverse trigonometric functions

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes differentiate thanks

OpenStudy (anonymous):

sin(u)= cos(u)

OpenStudy (anonymous):

cosu=sinu

OpenStudy (anonymous):

@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.

OpenStudy (anonymous):

oh thanks :)

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

*I meant good luck

OpenStudy (anonymous):

Thanks @CarlosGP :P

OpenStudy (anonymous):

thanks @CarlosGP :)))

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