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]}