what is the derivative of sqrt 2u + 1; u = (3v+9)^2; v= 2x - 7?

Question

with respect to x....

Answer 1

I'm sure you must be talking about an application of the chain rule, right?

f(x) = √(2(3(2x - 7) + 9)²) + 1)

f(u) = √(2u + 1)
u(v) = (3v + 9)²
v(x) = 2x - 7

Right?

It's unclear if you mean f(u) to √(2u + 1) or √2u + 1, but I think it doesn't make a difference in the end.

Then:

f(x) = f(u(v(x))), so by the chain rule:

f'(x) = f'(u(v(x))) × u'(v(x)) × v'(x)

= 1/2 × 1/√(2(3(2x - 7) + 9)² + 1) × 2 × // f'(u(v(x))
2 × (3(2x - 7) + 9) × 3 × // u'(v(x))
2 // v'(x)

Combining and simplifying:

12/√(2((6x - 21) + 9)² + 1) × ((6x - 21) + 9) =

12/√(2(6x - 12)² + 1) × (6x - 12) =

12(6x - 12)/√(2(36x² -144x + 144) + 1) =

72(x - 2)/√(72x² -288x + 289)

Very messy...

Answer 2

The derivative with respect to what variable, u, v, x or something else?

Bah.. let's assume x.

d/dx (√2u + 1) = [(1/2)(2u)/√2u ] * du/dx

du/dx = 2(3v+9) * 3v * dv/dx

dv/dx = 2

d/dx (√2u + 1) = 2*2*3*3*v*(v+3)*u/√2u

Now you have to substitute back in for v and u to get it all in terms of x if you so choose.

Answer 3

2u'/(2 sqrt(2u+1)

u'/sqrt(2u+1)

u'=2.3.v'. (3v+9)

u'=6.2(6x-21+9)
=72(x-2)

Therefore
72(x-2) / sqrt(2(6x-21+9)^2)+1)

72(x-2) / sqrt(2(6x-12)^2)+1)

Answer 4

2u+1=2(6x-12)^2 +1=72x^2 -288x +289
so the derivative is: (72*2)x-288=144x-288

Answer 5

For (2u)^½ + 1
δy/δu = ½(2u)^-½ + 0
δy/δu = ½(2u)^-½
δy/δu = 1/ [2(2u)^½]
δy/δu = 1/ [2√(2u)]


For u = (3v + 9)²:
u = (3v + 9)²
δu/δv = 2(3v + 9).3
δu/δv = 6(3v + 9)
δu/δv = 18v + 54

For v = 2x - 7:
v = 2x - 7
δv/δx = 2

Answer 6

The derivative of f(x) = sqrt[ 2 * u(x) ] + 1 w.r.t x

derivative[f(x)] = (+ or - )| 6 * sqrt(2) |

Answer 7

u = [3(2x-7) +9]^2 = (6x-21+9)^2 = (6x-12)^2
So you want derivative of 2(6x-12)^2 + 1
= 2*2*6(6x-12) = 144x-288