Dec 13, 2015 · Thus -x <= xcos (1/x) <= x, and so as x->0, xcos (1/x)->0. Then there is no asymptote at x=0, but xcos (1/x) is undefined at that point, meaning f (x) has a hole at x=0.

Explanation: #f (x)=x/ (sinxcosx)=g (x)/ (h (x))# #g (x)=x# #g' (x)=1# #h (x)=sinxcosx# #h' (x)=cosxcosx-sinxsinx=cos^2x-sin^2x# Using the double angle formula, #h ...

lim_ (x->0^+) (cot (x)-1)/x = oo lim_ (x->0^+) (cot (x) - 1/x) = 0 Try and learn how to input maths properly, it saves a lot of ambiguity. I don't know if you mean ...

LHS= ( cosx + sin^3 x- sinx)/ sin x = ( cosx - sin x (1- sin^2x))/ sin x = ( cosx - sin xcos^2x)/ sin x = cosx/sinx - (sin xcos^2x)/ sin x = cotx - cos^2x=RHS

Can you solve this limit simply by l'hopitle rule or is there a more complex method by which you can go about this?

Dec 17, 2017 · Develop the left side by using the trig identity: sin 2x -= 2sin x*cos x L.S = 2sin xcos x (sin x/cos x+ cos x/sin x) = 2sin xcos x* ((cos^2 x + sin^2 x))/(sin xcos x ) = 2(cos^2 x + sin^2 x) = 2 …

Feb 22, 2018 · f' (x)= 1+ (x-sinxcosx)/ (x^2cos^2x) Assuming f (x) =x +sinx/ (xcosx) f' (x) = 1 + d/dx (sinx/ (xcosx)) Apply quotient rule. f' (x) = 1+ (xcosx*cosx - sinx* d/dx ...

f^ (') (x) = (cos (x)ln (x)+sin (x)/x)/2 The function given is: =>f (x) = ln (sqrt (x^ (sin (x)))) Let's first simplify sqrt (x^sin (x)). Using the law: u (x) = e^ln ...

Feb 25, 2018 · Explanation: Differentiate both sides with respect to #x#. This means that every time we differentiate a term containing #y#, we should end up with an instance of #dy/dx#.

What is the equation of the line tangent to f (x) = 2x cos 3x at x = π 3?