How do you find an equation of the tangent line to the curve at the given point #y=tanx# and #(pi/4,1)#?
1 Answer
Aug 16, 2017
Explanation:
#•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"#
#dy/dx=sec^2x#
#x=pi/4tody/dx=1/cos^2(pi/4)=1/(1/sqrt2)^2=2#
#m_(color(red)"tangent")=2" and "(x_1,y_1)=(pi/4,1)#
#rArry-1=2(x-pi/4)#
#rArry=2x-pi/2+1larrcolor(red)" in slope-intercept form"#
