What is the derivative of sqrt(t^2+9) ?

The derivative of sqrt(t^2+9) is t/(sqrt(t^2+9))

What is the first derivative of sqrt(t^2+9) ?

The first derivative of sqrt(t^2+9) is t/(sqrt(t^2+9))

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derivative of

t^{2} + 9

\frac{t}{t ^{2} + 9}

Solution steps

\frac{d}{d t} (t^{2} + 9)

Apply the chain rule:

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

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

\frac{d}{d t} (t^{2} + 9) = 2 t

= \frac{1}{2 t ^{2} + 9} \cdot 2 t

Simplify

\frac{1}{2 t ^{2} + 9} \cdot 2 t : \frac{t}{t ^{2} + 9}

= \frac{t}{t ^{2} + 9}

x \to 2 lim (e^{x})

\int \frac{1}{6 - 36 x ^{2}} d x

\frac{2 - sec ( x )}{tan ( x )}

\int e^{x} 15 + e^{x} d x

\frac{d}{d x} (\frac{3}{8} (x^{\frac{4}{3}} - 2 x^{\frac{2}{3}}))

What is the derivative of sqrt(t^2+9) ?
The derivative of sqrt(t^2+9) is t/(sqrt(t^2+9))
What is the first derivative of sqrt(t^2+9) ?
The first derivative of sqrt(t^2+9) is t/(sqrt(t^2+9))