How do you differentiate #f(x) = x^3/(xcosx-2)# using the quotient rule?

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Answer 1 · 2016-01-08T02:45:33

#f '(x) = (x^4 *sin x + 2 x^3 *cos x - 6 x^2)/(x*cos x - 2)^2#

the quotient rule formula #d/dx(u/v)= (v*(du)/dx-u*(dv)/dx)/v^2#

the given: differentiate #f(x) = x^3/(x * cos x - 2)#

#f '(x) = (( x*cos x - 2)(3x^2) - x^3*(x*(-sin x)+(cos x)*(1)-0))/(x*cos x - 2)^2#

#f '(x)=(3x^3*cos x-6x^2+x^4*sin x-x^3*cos x+0)/(x*cos x-2)^2#

#f '(x)=((3x^3-x^3)*cos x +x^4*sin x-6x^2)/(x*cos x-2)^2#

#f '(x)=(2x^3*cos x+x^4*sin x-6x^2)/(x*cos x-2)^2#

the final simplification

#f '(x)=(x^4*sin x+ 2x^3*cos x-6x^2)/(x*cos x-2)^2#