What is the ((100)/(ln(2x+1)))^' ?

The ((100)/(ln(2x+1)))^' is -(200)/(ln^2(2x+1)(2x+1))

What is the first ((100)/(ln(2x+1)))^' ?

The first ((100)/(ln(2x+1)))^' is -(200)/(ln^2(2x+1)(2x+1))

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(\frac{100}{ln ( 2 x + 1 )})^{^{'}}

- \frac{200}{ln ^{2} ( 2 x + 1 ) ( 2 x + 1 )}

Solution steps

(\frac{100}{ln ( 2 x + 1 )})^{'}

Take the constant out:

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

= 100 (\frac{1}{ln ( 2 x + 1 )})^{^{'}}

Apply exponent rule:

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

= 100 (ln^{- 1} (2 x + 1))^{^{'}}

Apply the chain rule:

- \frac{1}{( ln ( 2 x + 1 ) ) ^{2}} (ln (2 x + 1))^{'}

= - \frac{1}{( ln ( 2 x + 1 ) ) ^{2}} (ln (2 x + 1))^{'}

(ln (2 x + 1))^{'} = \frac{2}{2 x + 1}

= 100 (- \frac{1}{( ln ( 2 x + 1 ) ) ^{2}} \cdot \frac{2}{2 x + 1})

Simplify

100 (- \frac{1}{ln ^{2} ( 2 x + 1 )} \cdot \frac{2}{2 x + 1}) : - \frac{200}{ln ^{2} ( 2 x + 1 ) ( 2 x + 1 )}

= - \frac{200}{ln ^{2} ( 2 x + 1 ) ( 2 x + 1 )}

\int 2 y^{2} ln (y) d y

7 e^{x^{2}}

\int \frac{tan ^{2} ( x )}{sin ( x )} d x

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

\frac{d y}{d x} = 2 x 1 - y^{2}

What is the ((100)/(ln(2x+1)))^' ?
The ((100)/(ln(2x+1)))^' is -(200)/(ln^2(2x+1)(2x+1))
What is the first ((100)/(ln(2x+1)))^' ?
The first ((100)/(ln(2x+1)))^' is -(200)/(ln^2(2x+1)(2x+1))