What is the equation of the line tangent to #f(x)=1/(1-3x)^2 # at #x=2#?
1 Answer
Apr 13, 2017
Explanation:
The equation of the tangent in
#color(blue)"point-slope form"# is.
#• y-y_1=m(x-x_1)#
#"where "color(red)( m=f'(2))" and " (x_1,y_1)" a point on the line"#
#f(x)=1/(1-3x)^2=(1-3x)^-2# differentiate using the
#color(blue)"chain rule"#
#"Given " f(x)=g(h(x))" then"#
#• f'(x)=g'(h(x))xxh'(x)#
#rArrf'(x)=-2(1-3x)^-3xxd/dx(1-3x)#
#color(white)(rArrf'(x))=6(1-3x)^-3=6/(1-3x)^3#
#rArrf'(2)=6/(-5)^3=-6/125=m_("tangent")#
#"and " f(2)=1/(-5)^2=1/25to(2,1/25)#
#"Using " m=-6/125" and " (x_1,y_1)=(2,1/25)#
#"then " y-1/25=-6/125(x-2)#
#rArry=-6/125x+17/125#