Find dy/dx and the gradient of the curve at the given value of y: y=(4x-5)ˆ3, y=27

Question

jack1 · Answer

i only know how to do this longhand, hope that's ok?

anonymous · Answer

Yup it's okay

jack1 · Answer

k, so y=(4x-5)ˆ3
            =(4x-5)(4x-5)(4x-5)
            = 64x^3   - 240x^2    + 300x   -125

anonymous · Answer

Is there any shorter method instead of having to expand it out?

jack1 · Answer

that's the thing, i only know how to do derivatives longhand (expanding first)
so dunno, sorry

.sam. · Answer

Just use power and chain rule

jack1 · Answer

y  = 64x^3   - 240x^2    + 300x   -125
so dy/dx = (64*3)x^2  -  (240*2)x   + 300

.sam. · Answer

$$y=(4x-5)^3 \ \ y'=3(4x-5)^2[\frac{d}{dx}[4x-5]]$$

campbell_st · Answer

well use the chain rule to differentiate 

$$y' = 3(4x - 5)^2 \times 4$$

so the derivative is 

$$y' = 12(4x - 5)^2$$

to find the gradient at y = 27 you need to find the value of x that makes$$27 = (4x - 5)^3$$ 
 then substitute it into the derivative

anonymous · Answer

When finding x, is the only way to expand it out? Or is there a shorter method?

campbell_st · Answer

nope take the cube root of both sides so 

3 = 4x - 5

solve for x

anonymous · Answer

@campbell_st Thanks!!