# Find the equation of tangent line to the function y=-1/x^3 at the pointx=1. write your answer in the gradient form, y=mx+c

## Find the equation of tangent line to the function y=-1/x^3 at the pointx=1.

Step-by-step solution-:

The given function is:

y=-\frac{1}{x^{3}}

We know that the first derivative of a function at any certain point (x,y) is equal to the slope of the tangent line.

Calculate the value of y for x=1:

y=-\frac{1}{(1)^{3}}

After simplifying we get,

y=-1

Let be m is the slope of the tangent line:

m= First derivative of y at (1,-1):

It means,

m=\frac{dy}{dx}

Find the first derivative of y:

y'=\frac{d(-\frac{1}{x^3})}{dx}=\frac{3}{x^4}

Substitute the value of x=1 into the above expression:

y'=\frac{dy}{dx}=\frac{3}{(1)^4}=3

It means,

m=3

We know that equation of a tangent line is:

(y-y_{1})=m\cdot(x-x_{1})

Substitute the point (1, -1) and slope m=3 into Point-Slope Form:

(y-(-1))=3\cdot(x-1)

After simplifying we get

y+1=3x-3=>y=3x-4
y=3x-4:

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