find the derivative of y=sinx(x+cscx)

Question

jhannybean · Answer

Use the product rule here. $f'\cdot g +g'\cdot f$

jhannybean · Answer

Your functions: $$f=\sin(x)$$$$g=x+\csc(x)$$

jhannybean · Answer

$$y=\color{red}{\sin(x)}\color{blue}{(x+\csc(x))}$$

anonymous · Answer

this is what i got cosx(x+cscx)+sinx(1-cscxcotx) is that right, how would you simplify that

jhannybean · Answer

Let's check

jhannybean · Answer

$$f'=\cos(x)$$$$g'=1-\csc(x)\cot(x)$$

jhannybean · Answer

Therefore with the product rule: $$y' = \color{red}{\cos(x)}(x+\csc(x)) +\color{blue}{(1-\csc(x)\cot(x))}(\sin(x))$$

anonymous · Answer

sin(x + csc x) will be chain rule as (x + csc x) is considered as x

here we have sin x(x + csc x)

So all you need to do is use the UV rule that is.. is is u'v + uv'

dy/dx = d/dx(sin x) x (x + csc x) + sinx x d/dx(x + csc x)

= cos x(x + csc x) + sinx(1 + sec^2x)

Can also write this as.. xcos x + cosxcosecx + sinx + sinxsec^2x

hmm play around with it maybe.. 

cosec x can be written as 1/sin

xcos x + cosx x 1/sin x + sinx + sinxsec^2x

recall

cos x/sin x = cot x

xcos x + cotx + sinx + sinxsec^2x = dy/dx

anonymous · Answer

ah ok thanks