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

The derivative of #y# will be the sum of the derivatives of #x^2# and #cos^-1x#. We will find these separately.

DERIVATIVE OF #bb(x^2)#

Since you're expected to find the derivative of #cos^-1x#, it's likely you already know how to differentiate #x^2#. In case you've forgotten, you'll need the power rule.

The power rule states that the derivative of #x^n# is equal to #d/dxx^n=nx^(n-1)#.

So, for #x^2#, we see that the derivative of #x^2# is #d/dxx^2=2x^(2-1)=2x^1=2x#.

DERIVATIVE OF #bb(cos^-1x)#

For this, we will need to do some manipulation. First, let:

#z=cos^-1x#

By the definition of the inverse trig functions (or inverse functions in general) this tells us that

#cos(z)=x#

We now should take the derivative of both sides (with respect to #x#). On the left-hand side, this will require the chain rule since #z# is its own function.

#d/dxcos(z)=d/dxx#

#-sin(z)*(dz)/dx=1#

We then should solve for #dz/dx#, which is the derivative of #cos^-1x#.

#dz/dx=-1/sin(z)#

We can rewrite this in terms of our original function. Remember, #cos(z)=x#. Furthermore, from the Pythagorean identity #sin^2(z)+cos^2(z)=1# we can say that #sin(z)=sqrt(1-cos^2(z))#.

#d/dxcos^-1x=-1/sqrt(1-cos^2(z))#

And since #cos(z)=x#, we can replace #cos^2(z)# with #x^2#:

#d/dxcos^-1x=-1/sqrt(1-x^2)#

PUTTING THEM TOGETHER

We then see that:

#dy/dx=d/dxx^2+d/dxcos^-1x#

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

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