What is the derivative of y=8sqrt(9x^2+7) ?

The derivative of y=8sqrt(9x^2+7) is (72x)/(sqrt(9x^2+7))

What is the first derivative of y=8sqrt(9x^2+7) ?

The first derivative of y=8sqrt(9x^2+7) is (72x)/(sqrt(9x^2+7))

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derivative of

y = 8 9 x^{2} + 7

\frac{72 x}{9 x ^{2} + 7}

Solution steps

\frac{d}{d x} (8 9 x^{2} + 7)

Take the constant out:

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

= 8 \frac{d}{d x} (9 x^{2} + 7)

Apply the chain rule:

\frac{1}{2 9 x ^{2} + 7} \frac{d}{d x} (9 x^{2} + 7)

= \frac{1}{2 9 x ^{2} + 7} \frac{d}{d x} (9 x^{2} + 7)

\frac{d}{d x} (9 x^{2} + 7) = 18 x

= 8 \cdot \frac{1}{2 9 x ^{2} + 7} \cdot 18 x

Simplify

8 \cdot \frac{1}{2 9 x ^{2} + 7} \cdot 18 x : \frac{72 x}{9 x ^{2} + 7}

= \frac{72 x}{9 x ^{2} + 7}

y = arccos (4 x)

x \to \infty lim (x + 1)

(e^{t} cos (t) + e^{t} sin (t) + e^{t})^{^{'}}

cos^{2} (x)

\int 5^{x} ln (5) d x

What is the derivative of y=8sqrt(9x^2+7) ?
The derivative of y=8sqrt(9x^2+7) is (72x)/(sqrt(9x^2+7))
What is the first derivative of y=8sqrt(9x^2+7) ?
The first derivative of y=8sqrt(9x^2+7) is (72x)/(sqrt(9x^2+7))