did it work?
See your previous post for a clue, and read up on inverse function derivatives.
suppose f(x) = (4/5)x^2 + cosx for x≥0 and g is the inverse of f. find g'(1.5) to the nearest thousandth.
would I just swap x and y and then find the derivative of that equation to find g'(x) and then plug in g'(1.5)? When I swapped x and y I got x=(4/5)y^2 + cosy and then do I do implicit differentiation? I at implicit differentiation so I'm stuck already :(
(i posted it incorrectly before my bad)
See your previous post for a clue, and read up on inverse function derivatives.
I get f(1.18) = 1.5
so g'(1.5) = 1/f'(1.18)
How did you get 1.18 from there?
1. Swap x and y in the equation f(x) = (4/5)x^2 + cos(x), which gives you x = (4/5)y^2 + cos(y).
2. Implicitly differentiate both sides of the equation with respect to x. Remember that y is a function of x, so you need to apply the chain rule when differentiating y terms.
d/dx [x] = d/dx [(4/5)y^2 + cos(y)]
1 = (4/5)(2y)(dy/dx) - sin(y)(dy/dx).
3. Now, you need to solve for dy/dx. Start by isolating the dy/dx term:
(4/5)(2y)(dy/dx) - sin(y)(dy/dx) = 1.
(8/5)y(dy/dx) - sin(y)(dy/dx) = 1.
[(8/5)y - sin(y)](dy/dx) = 1.
4. Divide both sides of the equation by [(8/5)y - sin(y)] to solve for dy/dx:
dy/dx = 1 / [(8/5)y - sin(y)].
5. Finally, evaluate g'(1.5) by plugging in y = 1.5 into the expression for dy/dx:
g'(1.5) = 1 / [(8/5)(1.5) - sin(1.5)].
At this point, you can use a calculator to compute the value of g'(1.5) to the nearest thousandth.