find the equation of the tangent to y=x-(a^3/x^2) at the point (a,0)
how to i work this out??can someone please show me how to do it step by step?thanks!!!
Take the dervative of y
y= x- a^3 x^-2
y'=1+2a^3/x^3
so the equation for the tangent is
y= mx + b where m is y' above evaluated at (a,0)
solve for b at that tangent point.
y= x- a^3 x^-2
y'=1+2a^3/x^3
how do i work out the dervative of y?
I just did it for your, y' is the derivative.
Sorry! i mean how do you know the derivative of y= x- a^3 x^-2 is y'=1+2a^3/x^3?
i know how to change y=x-(a^3/x^2) to y= x- a^3 x^-2 but then i got lost.
To find the derivative of y = x - (a^3 / x^2), you can apply the power rule and the constant multiple rule. Here's how you can do it step by step:
Step 1: Rewrite the equation in the form y = x - (a^3 / x^2).
Step 2: Differentiate each term separately. The derivative of x with respect to x is 1. The derivative of (a^3 / x^2) with respect to x can be found using the power rule:
d/dx (a^3 / x^2) = -2(a^3 / x^3) (-1) = 2(a^3 / x^3).
Step 3: Combine the derivatives of each term:
dy/dx = d/dx (x - (a^3 / x^2)) = d/dx x - d/dx (a^3 / x^2) = 1 + 2(a^3 / x^3).
So, the derivative of y with respect to x is dy/dx = 1 + 2(a^3 / x^3).
Now that you have the derivative, you can proceed with finding the equation of the tangent line to the curve at the point (a, 0).