How do you find the slope of a tangent line to the graph of the function# f(x)= -5/x^3# at x=9?

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Answer 1 · 2016-11-27T00:58:19

#f'(9)=5/2187#

#f(x)=frac{-5}{x^3}#

We want to find #f'(9)#, which is the slope of the tangent line at #x=9#.

Rewritten with a negative exponent:
#f(x)=-5x^(-3)#

Use the power rule to differentiate:
#f'(x)=(-3)(-5)(x^(-4))#
#f'(x)=15x^(-4)#

#f'(x)=frac{15}{x^4}#

Plug in 9 to find #f'(9)#:
#f'(9)=frac{15}{(9)^4}#
#f'(9)=15/6561#

#f'(9)=5/2187#