What is the equation of the tangent to the curve  y=9tanx  at the point where x=(2pi)/3?

What is the equation of the tangent to the curve  y=9tanx  at the point where x=(2pi)/3?

ReportAnswer 1

$y = 36 x - 24 \pi - 9 \sqrt{3}$

Explanation:

$y = 9 \tan x$

Differentiating wrt $x$ gives:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 9 {\sec}^{2} x$

So, When $x = \frac{2 \pi}{3} \implies y = 9 \tan \left(\frac{2 \pi}{3}\right) = - 9 \sqrt{3}$
and, $\frac{\mathrm{dy}}{\mathrm{dx}} = 9 {\sec}^{2} \left(\frac{2 \pi}{3}\right) = 9 {\left(- 2\right)}^{2} = 36$

Thus, the tangent has gradient $m = 36$ and passes through $\left(\frac{2 \pi}{3} , - 9 \sqrt{3}\right)$

So, using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the required equation is:
 y-(-9sqrt(3))) = 36(x-(2pi)/3) 
$\therefore y + 9 \sqrt{3} = 36 x - \frac{72 \pi}{3}$
$\therefore y = 36 x - 24 \pi - 9 \sqrt{3}$

answered by: Steve M Shwetank Mauria
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