What is the derivative of sqrt(x^2-4)sqrt(x+3) ?

The derivative of sqrt(x^2-4)sqrt(x+3) is (3x^2+6x-4)/(2sqrt(x+3)\sqrt{x^2-4)}

What is the first derivative of sqrt(x^2-4)sqrt(x+3) ?

The first derivative of sqrt(x^2-4)sqrt(x+3) is (3x^2+6x-4)/(2sqrt(x+3)\sqrt{x^2-4)}

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\frac{d}{d x} (x^{2} - 4 x + 3)

\frac{3 x ^{2} + 6 x - 4}{2 x + 3 x ^{2} - 4}

Solution steps

\frac{d}{d x} (x^{2} - 4 x + 3)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} - 4, g = x + 3

= \frac{d}{d x} (x^{2} - 4) x + 3 + \frac{d}{d x} (x + 3) x^{2} - 4

\frac{d}{d x} (x^{2} - 4) = \frac{x}{x ^{2} - 4}

\frac{d}{d x} (x + 3) = \frac{1}{2 x + 3}

= \frac{x}{x ^{2} - 4} x + 3 + \frac{1}{2 x + 3} x^{2} - 4

Simplify

\frac{x}{x ^{2} - 4} x + 3 + \frac{1}{2 x + 3} x^{2} - 4 : \frac{3 x ^{2} + 6 x - 4}{2 x + 3 x ^{2} - 4}

= \frac{3 x ^{2} + 6 x - 4}{2 x + 3 x ^{2} - 4}

e^{- 4 s}

\frac{\partial}{\partial z} (- 9 cos (3 x + yz) z)

n = 1 \sum \infty n e^{- 5 n}

cos^{4} (θ + 9)

y = x^{4} + 4 e^{x}, (0, 4)

What is the derivative of sqrt(x^2-4)sqrt(x+3) ?
The derivative of sqrt(x^2-4)sqrt(x+3) is (3x^2+6x-4)/(2sqrt(x+3)\sqrt{x^2-4)}
What is the first derivative of sqrt(x^2-4)sqrt(x+3) ?
The first derivative of sqrt(x^2-4)sqrt(x+3) is (3x^2+6x-4)/(2sqrt(x+3)\sqrt{x^2-4)}