anonymous
  • anonymous
Find dy/dx and the gradient of the curve at the given value of y: y=(4x-5)ˆ3, y=27
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Jack1
  • Jack1
i only know how to do this longhand, hope that's ok?
anonymous
  • anonymous
Yup it's okay
Jack1
  • Jack1
k, so y=(4x-5)ˆ3 =(4x-5)(4x-5)(4x-5) = 64x^3 - 240x^2 + 300x -125

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anonymous
  • anonymous
Is there any shorter method instead of having to expand it out?
Jack1
  • Jack1
that's the thing, i only know how to do derivatives longhand (expanding first) so dunno, sorry
.Sam.
  • .Sam.
Just use power and chain rule
Jack1
  • Jack1
y = 64x^3 - 240x^2 + 300x -125 so dy/dx = (64*3)x^2 - (240*2)x + 300
.Sam.
  • .Sam.
\[y=(4x-5)^3 \\ \\ y'=3(4x-5)^2[\frac{d}{dx}[4x-5]]\]
campbell_st
  • campbell_st
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
  • anonymous
When finding x, is the only way to expand it out? Or is there a shorter method?
campbell_st
  • campbell_st
nope take the cube root of both sides so 3 = 4x - 5 solve for x
anonymous
  • anonymous
@campbell_st Thanks!!

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