# Find the slope of the tangent line to y=x^5-3x^-3 at x=-2

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The slope of a tangent line to a curve at any point is the value of the first derivative at that point.

The function we have is y = x^5 - 3x^(-3)

y' = 5x^4 - (-3*3)x^(-4)

=> y' = 5x^4 + 9x^-4

At x = -2

y' = 5*(-2)^4 + 9*(-2)^(-4)

=> 5*16 + 9/ 16

=> 1289/16

**The slope of the tangent to the curve y=x^5-3x^-3 at x=-2 is 1289/16**

The slope of a tangent drawn on a curve y = f(x) at a point where x = a is given by the f'(a) where f'(x) is the first derivative of y = f(x).

Here y = f(x) is y=x^5-3x^-3

f'(x) = (x^5 - 3x^-3)'

= 5x^4 - 3*(-3)*x^(-3-1)

= 5x^4 + 9*x^-4

At the point where x = -2

f'(-2) = 5*(-2)^4 + 9*(-2)^-4

= 80 + 9/16

= 1289/16

The slope of the required tangent is 1289/16