What is the derivative of 4t^{-1/8}+4ln(t)+ln(5) ?

The derivative of 4t^{-1/8}+4ln(t)+ln(5) is (8t^{9/8}-t)/(2t^{17/8)}

What is the first derivative of 4t^{-1/8}+4ln(t)+ln(5) ?

The first derivative of 4t^{-1/8}+4ln(t)+ln(5) is (8t^{9/8}-t)/(2t^{17/8)}

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derivative of

4 t^{- \frac{1}{8}} + 4 ln (t) + ln (5)

\frac{8 t ^{\frac{9}{8}} - t}{2 t ^{\frac{17}{8}}}

Solution steps

\frac{d}{d t} (4 t^{- \frac{1}{8}} + 4 ln (t) + ln (5))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (4 t^{- \frac{1}{8}}) + \frac{d}{d t} (4 ln (t)) + \frac{d}{d t} (ln (5))

\frac{d}{d t} (4 t^{- \frac{1}{8}}) = - \frac{1}{2 t ^{\frac{9}{8}}}

\frac{d}{d t} (4 ln (t)) = \frac{4}{t}

\frac{d}{d t} (ln (5)) = 0

= - \frac{1}{2 t ^{\frac{9}{8}}} + \frac{4}{t} + 0

Simplify

- \frac{1}{2 t ^{\frac{9}{8}}} + \frac{4}{t} + 0 : \frac{8 t ^{\frac{9}{8}} - t}{2 t ^{\frac{17}{8}}}

= \frac{8 t ^{\frac{9}{8}} - t}{2 t ^{\frac{17}{8}}}

\int (x^{2} + 3 x + 2) e^{x} d x

x \to 3 lim (4 x^{2} - \frac{1}{x})

\int \frac{2 x ^{3} + 7 x ^{2} + 2 x + 9}{2 x + 3} d x

x e^{- x^{10}}

y^{^{'}} - 2 y = - 4 y^{3}

What is the derivative of 4t^{-1/8}+4ln(t)+ln(5) ?
The derivative of 4t^{-1/8}+4ln(t)+ln(5) is (8t^{9/8}-t)/(2t^{17/8)}
What is the first derivative of 4t^{-1/8}+4ln(t)+ln(5) ?
The first derivative of 4t^{-1/8}+4ln(t)+ln(5) is (8t^{9/8}-t)/(2t^{17/8)}