How do you differentiate #y=2cos(x)+6cos^-1(x)#?

2 Answers
Douglas K.
Jun 17, 2017

Given: #y=2cos(x)+6cos^-1(x)#

Differentiate each term:

#dy/dx=(d(2cos(x)))/dx+(d(6cos^-1(x)))/dx#

Bring the constants outside:

#dy/dx=2(d(cos(x)))/dx+6(d(cos^-1(x)))/dx#

We know that #d/dxcos(x) = -sin(x)#

#dy/dx = -2sin(x) +6(d(cos^-1(x)))/dx #

We know that #d/dxcos^-1(x) = -1/sqrt(1-x^2)#:

#dy/dx = -2sin(x) -6/sqrt(1-x^2) #

Jim G.
Jun 17, 2017

#dy/dx=-2sinx-6/sqrt(1-x^2)#

Explanation:

#"using the "color(blue)"standard derivatives"#

#• d/dx(cosx)=-sinx#

#• d/dx(cos^-1x)=-1/(sqrt(1-x^2)#

#y=2cosx+6cos^-1x#

#rArrdy/dx=-2sinx-6/(sqrt(1-x^2)#

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
3142 views around the world
You can reuse this answer
Creative Commons License