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

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

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

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

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derivative of

2 t^{- \frac{1}{4}} + 2 ln (t) + ln (5)

\frac{4 t ^{\frac{5}{4}} - t}{2 t ^{\frac{9}{4}}}

Solution steps

\frac{d}{d t} (2 t^{- \frac{1}{4}} + 2 ln (t) + ln (5))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (2 t^{- \frac{1}{4}}) + \frac{d}{d t} (2 ln (t)) + \frac{d}{d t} (ln (5))

\frac{d}{d t} (2 t^{- \frac{1}{4}}) = - \frac{1}{2 t ^{\frac{5}{4}}}

\frac{d}{d t} (2 ln (t)) = \frac{2}{t}

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

= - \frac{1}{2 t ^{\frac{5}{4}}} + \frac{2}{t} + 0

Simplify

- \frac{1}{2 t ^{\frac{5}{4}}} + \frac{2}{t} + 0 : \frac{4 t ^{\frac{5}{4}} - t}{2 t ^{\frac{9}{4}}}

= \frac{4 t ^{\frac{5}{4}} - t}{2 t ^{\frac{9}{4}}}

y^{^{''}} - y^{^{'}} + 0.25 y = 0

\int e^{4 x} sin (b x) d x

f (x) = x + 9

\int 5 x^{\frac{3}{2}} d x

x \to \infty lim ((x) (e^{\frac{1}{x}}))

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