What is the equation of the tangent line of #y=1/x^2# at #x=1#?

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Answer 1 · 2016-07-08T20:59:40

To make it look easier, we can see that #y=(1)/(x^2)->y=x^(-2)#, so we can apply the power rule for derivatives, giving us

#y'=-2x^(-1)=(-2)/(x)#

#y'(1) = -2/1 = -2#

Knowing that #y(1) = 1#, we can use the point-slope formula to get an equation of the tangent line at #x = 1#, which is

#y - 1 = -2(x-1)# or

#y=-2x+3#

The power rule for derivatives states that

#d/dx[u^n] = n*u^(n-1)#.