What is the derivative of 3/(sqrt(6t+3)) ?

The derivative of 3/(sqrt(6t+3)) is -9/((6t+3)^{3/2)}

What is the first derivative of 3/(sqrt(6t+3)) ?

The first derivative of 3/(sqrt(6t+3)) is -9/((6t+3)^{3/2)}

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derivative of

\frac{3}{6 t + 3}

- \frac{9}{( 6 t + 3 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d t} (\frac{3}{6 t + 3})

Take the constant out:

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

= 3 \frac{d}{d t} (\frac{1}{6 t + 3})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 3 \frac{d}{d t} ((6 t + 3)^{- \frac{1}{2}})

Apply the chain rule:

- \frac{1}{2 ( 6 t + 3 ) ^{\frac{3}{2}}} \frac{d}{d t} (6 t + 3)

= - \frac{1}{2 ( 6 t + 3 ) ^{\frac{3}{2}}} \frac{d}{d t} (6 t + 3)

\frac{d}{d t} (6 t + 3) = 6

= 3 (- \frac{1}{2 ( 6 t + 3 ) ^{\frac{3}{2}}} \cdot 6)

Simplify

3 (- \frac{1}{2 ( 6 t + 3 ) ^{\frac{3}{2}}} \cdot 6) : - \frac{9}{( 6 t + 3 ) ^{\frac{3}{2}}}

= - \frac{9}{( 6 t + 3 ) ^{\frac{3}{2}}}

f^{^{'}} (t) = \frac{8 t}{8 + t}

f (x) = (2 x^{2} - 1)^{\frac{1}{2}}

y^{^{''}} + 8 y^{^{'}} + 65 y = 0, y (0) = 0, y^{^{'}} (0) = 0

x \to - 2 lim (x^{2} - 2 x^{2} + 2 x + 3)

\frac{d}{d x} (\frac{tan ( x ) - 1}{sec ( x )})

What is the derivative of 3/(sqrt(6t+3)) ?
The derivative of 3/(sqrt(6t+3)) is -9/((6t+3)^{3/2)}
What is the first derivative of 3/(sqrt(6t+3)) ?
The first derivative of 3/(sqrt(6t+3)) is -9/((6t+3)^{3/2)}