what is the derivative of" r*x*cosx + r Ln r sinx " with respect to x as a variable?

Answer 1

(r*x)*(-sin(x))+(cos(x))*(r)+
(0)*(lnrsin(x))+(r)*(cot(x))

for the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)

sorry it cut off on the top. hopefully this works

Answer 2

You have to differentiate wrt x. So, x is ur variable and all others are constants (here, r is a constant.)
So, the answer is
r*(cosx - x*sinx) + r Ln r * cosx.

Derivative of cosx is -sinx.
Derivative of sinx is cosx.
Derivative of x is 1.

Answer 3

is r a constant or a variable? also, how is the r Ln r sinx grouped? ARe they being multipled together of is it r * ln(r * sinx)?


actually it's (rx)(-sinx)+(r)(cosx) + r[(1/rsinx)(rcosx)]
r(cosx-x*sinx) + (r^2)(cotx)
r(cosx - xsinx + rcotx)

Answer 4

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Answer 5

Firstly, by considering the question as follows:
> r *x* cos(x) + r *ln(r *sin(x))
The answer is:
> r*cos(x)-r*x*sin(x)+r*cos(x)/sin(x)

Secondly, by combining r to r^2 like:
> r*x*cos(x)+r^2*ln(sin(x))
The answer is:
> r*cos(x)-r*x*sin(x)+r^2*cos(x)/sin(x)

You should choose the proper settings and see the answer. Anyway you just have two choices.

Answer 6

f(x)=r*x*cosx + r Ln r sinx
assume you mean this:
f(x)=r*x*cosx + r *(Ln (r sinx))
f'(x)=r*cosx - r*x*sinx + r*(1/(rsinx))*(r*cosx)
f'(x)=r*(cosx-x*sinx) + r*cotx
f'(x)=r*(cosx- x*sinx + cotx)