Question

Can helper help me? I have mistake

Answer 1

Your work seems fine to me. All that's left is simplification. Any of the following forms are equivalent.

$\dfrac{dy}{dx} = \dfrac{3b\sqrt{ax} - (3bx+ac) \left(\frac12 (ax)^{-1/2}\right)}{\left(\sqrt{ax}\right)^2}$

$\dfrac{dy}{dx} = \dfrac{3b\sqrt{ax} - \frac{3bx+ac}{2\sqrt{ax}}}{ax}$

$\dfrac{dy}{dx} = \dfrac{6abx - (3bx+ac)}{2(ax)^{3/2}}$

$\dfrac{dy}{dx} = \dfrac{(6ab-3b)x - ac}{2(ax)^{3/2}}$

$\dfrac{dy}{dx} = \dfrac{(6ab-3b)x}{2a^{3/2}x^{3/2}} - \dfrac{ac}{2a^{3/2}x^{3/2}}$

$\dfrac{dy}{dx} = \dfrac{6ab-3b}{2a^{3/2}x^{1/2}} - \dfrac c{2a^{1/2}x^{3/2}}$

Answer 2

By derivative rules, the first derivative of the function f(x) = (3 · b · x + a · c) / [√(a · x)] is equal to f'(x) = [(3 · b) · √(a · x) - (3 · b · x + a · c) · 0.5 ·√(a / x)] / (a · x).

How to use derivative rules for a division between two functions

Let f(x) a function of the form f(x) = g(x) / h(x), whose first derivative is defined by the following expression:

f'(x) = [g'(x) · h(x) - g(x) · h'(x)] / [h(x)]²      (1)

If we know that g(x) = 3 · b · x + a · c and h(x) = √(a · x), then the first derivative of the function is:

f(x) = (3 · b · x + a · c) / [√(a · x)]

g'(x) = 3 · b

h'(x) = 0.5 · a /√(a · x)

h'(x) = 0.5 ·√(a / x)

f'(x) = [(3 · b) · √(a · x) - (3 · b · x + a · c) · 0.5 ·√(a / x)] / [√(a · x)]²

f'(x) = [(3 · b) · √(a · x) - (3 · b · x + a · c) · 0.5 ·√(a / x)] / (a · x)

The first derivative of the function f(x) = (3 · b · x + a · c) / [√(a · x)] is equal to f'(x) = [(3 · b) · √(a · x) - (3 · b · x + a · c) · 0.5 ·√(a / x)] / (a · x).

To learn more on derivatives: brainly.com/question/25324584

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