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

The derivative of (1+5t)/(ln(t^2)) is (5tln(t^2)-2(1+5t))/(tln^2(t^2))

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

The first derivative of (1+5t)/(ln(t^2)) is (5tln(t^2)-2(1+5t))/(tln^2(t^2))

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d t} (1 + 5 t) = 5

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

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

Simplify

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

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

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

\int_{1}^{10} 5 x^{- 2} d x

\frac{d}{d x} (2 x^{3} - 5 x^{2} + 3 x - 1)

\frac{1}{s ( s ^{2} + 100 s + 10000 )}

\frac{\partial}{\partial z} (2 x y)

What is the derivative of (1+5t)/(ln(t^2)) ?
The derivative of (1+5t)/(ln(t^2)) is (5tln(t^2)-2(1+5t))/(tln^2(t^2))
What is the first derivative of (1+5t)/(ln(t^2)) ?
The first derivative of (1+5t)/(ln(t^2)) is (5tln(t^2)-2(1+5t))/(tln^2(t^2))