Yequals(9 x Superscript 4 Baseline minus 4 x squared plus 6 )Superscript 4 To find StartFraction dy Over dx EndFraction , write y as a function of u so that yequalsf(u) and uequalsg(x). What is uequalsg(x) in this case?

Step-by-step explanation:

Given the function

y = (9x⁴ — 4x² + 6)⁴

We need to find the derivative of y with respect to x i.e. dy/dx.

So let u = 9x⁴—4x² + 6

Then y = u²,

Then, y is a function of u, y=f(u)

Also, u is a function of x, u = g(x)

In this case,

u = g(x) = 9x⁴—4x² + 6

So let differentiate this function y(x).

This is a function of a function

Then, we need to find u'(x)

u (x) = 9x⁴—4x² + 6

Then, u'(x) = 36x³ — 8x

Also we need to find y'(u)

Then, y = u²

y'(u) = 2u

Using function of a function formula

dy / dx = dy/du × du/dx

y'(x) = y'(u) × u'(x)

y'(x) = 2u × 36x³ — 8x

y'(x) = 2u(36x³ — 8x)

Since, u = 9x⁴—4x² + 6

Therefore,

y'(t) = 2(9x⁴—4x² + 6)(36x³ — 8x)

So,

dy/dx = 2(9x⁴—4x² + 6)(36x³ — 8x)

dy/dx = (18x⁴—8x² + 12)(36x³ — 8x)

step-by-step explanation:

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step-by-step explanation: