What is the equation of the line tangent to #f(x)=1/(1-3x)^2 # at #x=2#?

1 Answer
Jim G.
Apr 13, 2017

#y=-6/125 x+17/125#

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#

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