d/dxsin^-1x=1/sqrt(1-x^2)ddxsin−1x=1√1−x2
d/dxcos^-1x=-1/sqrt(1-x^2)ddxcos−1x=−1√1−x2
tan^-1x=1/(1+x^2)tan−1x=11+x2
cot^-1x=-1/sqrt(1+x^2)cot−1x=−1√1+x2
sec^-1x=1/(xsqrt(x^2-1))sec−1x=1x√x2−1
csc^-1x=-1/(xsqrt(x^2-1))csc−1x=−1x√x2−1
One useful thing to notice is that the derivatives of all inverse "co" functions are equivalent to the derivatives of the original inverse function, but have a negative added.
Here's a proof for the derivative of the inverse sine function, if you want to avoid memorization. All the other ones can be proved in the same way.
y=sin^-1x hArr x=sinyy=sin−1x⇔x=siny, from the definition of an inverse function.
Differentiating x=siny:x=siny:
d/dx(x)=d/dx(siny)ddx(x)=ddx(siny)
1=cosy*dy/dx1=cosy⋅dydx (Implicit Differentiation)
dy/dx=1/cosydydx=1cosy
We need to get rid of cosy.cosy. Recall that we said x=sinyx=siny and recall the identity sin^2theta+cos^2theta=1sin2θ+cos2θ=1. This can be rewritten for yy and solved for cosine as follows:
cos^2y=1-sin^2ycos2y=1−sin2y
cosy=sqrt(1-sin^2y)cosy=√1−sin2y
Recalling that x=siny,x=siny, then sin^2y=x^2sin2y=x2
Thus,
cosy=sqrt(1-x^2)cosy=√1−x2
d/dxsin^-1x=1/sqrt(1-x^2)ddxsin−1x=1√1−x2
I recommend that you commit these integrals to memory- they will help you when you are learning the trig. substitution.
