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

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Answer 1 · 2016-10-23T19:56:19

$y = 36x-24pi -9sqrt(3)$

$y=9tanx$

Differentiating wrt $x$ gives:
$dy/dx =9sec^2x$

So, When $x=(2pi)/3 => y=9tan((2pi)/3) =-9sqrt(3)$
and, $dy/dx = 9sec^2((2pi)/3)=9(-2)^2=36$

Thus, the tangent has gradient $m=36$ and passes through $((2pi)/3,-9sqrt(3))$

So, using $y-y_1=m(x-x_1)$ the required equation is:
$y-(-9sqrt(3))) = 36(x-(2pi)/3)$
$:. y+9sqrt(3) = 36x-(72pi)/3$
$:. y = 36x-24pi -9sqrt(3)$

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