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

Explanation:

# 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)#

answered by: Steve M Shwetank Mauria
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What is the equation of the tangent to the curve # y=9tanx # at the point where #x=(2pi)/3#?
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