How do you find an equation of the tangent line to the curve at the given point $y=tan(4x^2)$ and $x=sqrtpi$ ?

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Answer 1 · 2017-06-18T16:09:27

$y=8sqrt(pi)x-8pi$

When $x=sqrt(pi)$ , then

$y=tan(4(sqrt(pi))^2)$

$y=tan(4pi)$

$y=0$

So we have the point $(sqrt(pi),0)$ . The slope, $m$ , is found by the derivative at this point.

The derivative of $tan(u)$ is $sec^2(u)(du)/dx$

$d/dx tan(4x^2)=sec^2(4x^2)xx8x$

When $x=sqrt(pi)$ , the slope is

$m=sec^2(4(sqrt(pi))^2)xx8(sqrt(pi))$

$m=(8sqrt(pi))/cos^2(4pi)$

$m=(8sqrt(pi))/1^2~~14.18$

You can use the point and the slope in the point-slope form of a line

$y-y_1=m(x-x_1)$

$y-0=8sqrt(pi)(x-sqrt(pi))$

$y=8sqrt(pi)x-8pi$

Here is a graph of the tangent function with the tangent line passing through the point $(sqrt(pi),0)$

graph{(y-tan(4x^2))(y-8sqrt(pi)x+8pi)=0[1.2,2.2,-3,3]}

How do you find an equation of the tangent line to the curve at the given point y=tan(4x^2) y=tan(4x^2) and x=sqrtpix=sqrtpi?