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

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Answer 1 · 2017-06-18T16:09:27

$y = 8 \sqrt{π} x - 8 π$ $y=8sqrt(pi)x-8pi$

When $x = \sqrt{π}$ $x=sqrt(pi)$ , then

$y = tan (4 {(\sqrt{π})}^{2})$ $y=tan(4(sqrt(pi))^2)$

$y = tan (4 π)$ $y=tan(4pi)$

$y = 0$ $y=0$

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

The derivative of $tan (u)$ $tan(u)$ is ${sec}^{2} (u) \frac{d u}{d x}$ $sec^2(u)(du)/dx$

$\frac{d}{d x} tan (4 x^{2}) = {sec}^{2} (4 x^{2}) \times 8 x$ $d/dx tan(4x^2)=sec^2(4x^2)xx8x$

When $x = \sqrt{π}$ $x=sqrt(pi)$ , the slope is

$m = {sec}^{2} (4 {(\sqrt{π})}^{2}) \times 8 (\sqrt{π})$ $m=sec^2(4(sqrt(pi))^2)xx8(sqrt(pi))$

$m = \frac{8 \sqrt{π}}{{cos}^{2} (4 π)}$ $m=(8sqrt(pi))/cos^2(4pi)$

$m = \frac{8 \sqrt{π}}{1^{2}} \approx 14.18$ $m=(8sqrt(pi))/1^2~~14.18$

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

$y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$

$y - 0 = 8 \sqrt{π} (x - \sqrt{π})$ $y-0=8sqrt(pi)(x-sqrt(pi))$

$y = 8 \sqrt{π} x - 8 π$ $y=8sqrt(pi)x-8pi$

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

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(4x2)y=tan(4x^2) and x=√πx=sqrtpi?