Can anyone explain the derrivative of anti-trig functions (especially arctan)?

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Answer 1 · 2018-04-06T19:41:15

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#y=arctanx#

The meaning of this inverse tangent function is:

#color(purple)("y is an angle whose tangent is x")#

Therefore, we can write it as:

#tany=x#

Now, we can take derivatives of both sides:

#sec^2ydy=dx#

But we have an identity that gives us:

#sec^2y=tan^2y+1#

Let's plug this in:

#(tan^2y+1)dy=dx#

Let's divide both sides by #dx#:

#(tan^2y+1)dy/dx=1#

Let's divide both sides by#(tan^2y+1)# to solve for #dy/dx#:

#dy/dx=1/(tan^2y+1)#

But because #tany=x#, we can say:

#tan^2y=x^2#

If we plug this into the #dy/dx# equation above we get:

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

Since we started with #y=arctanx#, we can plug it in:

#(d(arctanx))/dx=1/(x^2+1)#

You can use the same process for all the other inverse trigonometric functions and calculate their derivatives.