What is the derivative of f(x)=-4sqrt(6x^2-5x+10) ?

The derivative of f(x)=-4sqrt(6x^2-5x+10) is -(2(12x-5))/(sqrt(6x^2-5x+10))

What is the first derivative of f(x)=-4sqrt(6x^2-5x+10) ?

The first derivative of f(x)=-4sqrt(6x^2-5x+10) is -(2(12x-5))/(sqrt(6x^2-5x+10))

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derivative of

f (x) = - 4 6 x^{2} - 5 x + 10

- \frac{2 ( 12 x - 5 )}{6 x ^{2} - 5 x + 10}

Solution steps

\frac{d}{d x} (- 4 6 x^{2} - 5 x + 10)

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 4 \frac{d}{d x} (6 x^{2} - 5 x + 10)

Apply the chain rule:

\frac{1}{2 6 x ^{2} - 5 x + 10} \frac{d}{d x} (6 x^{2} - 5 x + 10)

= \frac{1}{2 6 x ^{2} - 5 x + 10} \frac{d}{d x} (6 x^{2} - 5 x + 10)

\frac{d}{d x} (6 x^{2} - 5 x + 10) = 12 x - 5

= - 4 \cdot \frac{1}{2 6 x ^{2} - 5 x + 10} (12 x - 5)

Simplify

- 4 \cdot \frac{1}{2 6 x ^{2} - 5 x + 10} (12 x - 5) : - \frac{2 ( 12 x - 5 )}{6 x ^{2} - 5 x + 10}

= - \frac{2 ( 12 x - 5 )}{6 x ^{2} - 5 x + 10}

x \to 0 lim (3 - x)

\frac{\partial}{\partial θ} (a cos (θ))

\frac{\partial}{\partial x} (5 x^{2} y + 6 x y^{3})

y^{^{'}} = 9 y^{2} sin (x)

f (x) = 3 x^{3} + 5 x^{2} + 5 x, at x = - 1

What is the derivative of f(x)=-4sqrt(6x^2-5x+10) ?
The derivative of f(x)=-4sqrt(6x^2-5x+10) is -(2(12x-5))/(sqrt(6x^2-5x+10))
What is the first derivative of f(x)=-4sqrt(6x^2-5x+10) ?
The first derivative of f(x)=-4sqrt(6x^2-5x+10) is -(2(12x-5))/(sqrt(6x^2-5x+10))