The equation of the tangent line is
In this section we compute derivatives involving and .
We begin by computing the derivative of the inverse trigonometric function . The following Pythagorean trigonometric identity will be needed:
This identity follows from by dividing both sides by .
We begin the derivation by using the fact that and are inverse functions, so that:
We differentiate both sides of this equation with respect to :
Using the Chain Rule on the left side gives:
Now, we can solve this for the derivative of :
Next, we use the Pythagorean Identity:
Finally, using the property of inverse functions:
Similar arguments can show that:
Find the equation of the tangent line to the graph of at
The point of tangency is since . The slope of the tangent line is given by . To
create the equation of the line we use the Point slope form: and we get or
.
Find the equation of the tangent line to the graph of at
The equation of the tangent line is
Find the equation of the tangent line to the graph of at
The equation of the tangent line is
We write as a product: with To use the product rule we need the derivatives: We can now write the derivative: \begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 2x\tan ^{-1}(x) + (1+x^2)\frac{1}{1+x^2} \\ &= 2x\tan ^{-1}(x) + 1. \end{align*}
The Chain Rule versions of these formulas are:
and
The equation of the tangent line is