Question

y = (6x⁴ – 5x³ + 2x + 3)⁶, find `("d"y)/("d"x)`

Solution: Given,

y = (6x⁴ – 5x³ + 2x + 3)⁶ 

Let u = `[6x^4 - 5x^3 + square + 3]`

∴ y = `"u"^square`

∴ `("d"y)/"du"` = 6u^6–1

∴ `("d"y)/"du"` = 6(  )⁵ 

and `"du"/("d"x) = 24x^3 - 15(square) + 2`

By chain rule,

`("d"y)/("d"x) = ("d"y)/square xx square/("d"x)`

∴ `("d"y)/("d"x) = 6(6x^4 - 5x^3 + 2x + 3)^square xx (24x^3 - 15x^2 + square)`

Answer 1

Given,

y = (6x⁴ – 5x³ + 2x + 3)⁶ 

Let u = [6x⁴ – 5x³ + 2x + 3]

∴ y = `"u"^6`

∴ `("d"y)/"du"` = 6u^6–1

∴ `("d"y)/"du"` = 6(u)⁵ 

and `"du"/("d"x) = 24x^3 - 15x^2 + 2`

By chain rule,

`("d"y)/("d"x) = ("d"y)/"du" xx "du"/("d"x)`

∴ `("d"y)/("d"x) = 6(6x^4 - 5x^3 + 2x + 3)^5 xx (24x^3 - 15x^2 + 2)`