What is the derivative of (10log_{4}(t))/t ?

The derivative of (10log_{4}(t))/t is (5(-2ln(2)log_{4}(t)+1))/(ln(2)t^2)

What is the first derivative of (10log_{4}(t))/t ?

The first derivative of (10log_{4}(t))/t is (5(-2ln(2)log_{4}(t)+1))/(ln(2)t^2)

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derivative of

\frac{10 lo g _{4} ( t )}{t}

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

Solution steps

\frac{d}{d t} (\frac{10 lo g _{4} ( t )}{t})

Take the constant out:

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

= 10 \frac{d}{d t} (\frac{lo g _{4} ( t )}{t})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 10 \cdot \frac{\frac{d}{d t} ( lo g _{4} ( t ) ) t - \frac{d t}{d t} lo g _{4} ( t )}{t ^{2}}

\frac{d}{d t} (lo g_{4} (t)) = \frac{1}{2 ln ( 2 ) t}

\frac{d t}{d t} = 1

= 10 \cdot \frac{\frac{1}{2 l n ( 2 ) t} t - 1 \cdot lo g _{4} ( t )}{t ^{2}}

Simplify

10 \cdot \frac{\frac{1}{2 l n ( 2 ) t} t - 1 \cdot lo g _{4} ( t )}{t ^{2}} : \frac{5 ( - 2 ln ( 2 ) lo g _{4} ( t ) + 1 )}{ln ( 2 ) t ^{2}}

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

\frac{\partial}{\partial x} (x^{2} sin (x y))

n = 0 \sum \infty 9^{n} x^{n}

9 x^{2} y^{^{'}} = y^{^{'}} + 9 x e^{- y}

\int - 2 x 8 - x^{2} d x

x \to 0 + lim (\frac{x - ∣ x ∣}{x})

What is the derivative of (10log_{4}(t))/t ?
The derivative of (10log_{4}(t))/t is (5(-2ln(2)log_{4}(t)+1))/(ln(2)t^2)
What is the first derivative of (10log_{4}(t))/t ?
The first derivative of (10log_{4}(t))/t is (5(-2ln(2)log_{4}(t)+1))/(ln(2)t^2)