What is the derivative of log_{2}(7x^2+x-2) ?

The derivative of log_{2}(7x^2+x-2) is (14x+1)/(ln(2)(7x^2+x-2))

What is the first derivative of log_{2}(7x^2+x-2) ?

The first derivative of log_{2}(7x^2+x-2) is (14x+1)/(ln(2)(7x^2+x-2))

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\frac{d}{d x} (lo g_{2} (7 x^{2} + x - 2))

\frac{14 x + 1}{ln ( 2 ) ( 7 x ^{2} + x - 2 )}

Solution steps

\frac{d}{d x} (lo g_{2} (7 x^{2} + x - 2))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( 7 x ^{2} + x - 2 )}{ln ( 2 )})

Take the constant out:

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

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (7 x^{2} + x - 2))

Apply the chain rule:

\frac{1}{7 x ^{2} + x - 2} \frac{d}{d x} (7 x^{2} + x - 2)

= \frac{1}{7 x ^{2} + x - 2} \frac{d}{d x} (7 x^{2} + x - 2)

\frac{d}{d x} (7 x^{2} + x - 2) = 14 x + 1

= \frac{1}{ln ( 2 )} \cdot \frac{1}{7 x ^{2} + x - 2} (14 x + 1)

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{7 x ^{2} + x - 2} (14 x + 1) : \frac{14 x + 1}{ln ( 2 ) ( 7 x ^{2} + x - 2 )}

= \frac{14 x + 1}{ln ( 2 ) ( 7 x ^{2} + x - 2 )}

\frac{d}{d x} (12 x - x^{2})

x \to 1 lim (\frac{( x + 1 )}{x - 1})

\frac{d}{d x} (x^{2} - \frac{1}{8} ln (x))

\int e^{- 2 t} t d t

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{( e ^{- 3 x} )}{x}

What is the derivative of log_{2}(7x^2+x-2) ?
The derivative of log_{2}(7x^2+x-2) is (14x+1)/(ln(2)(7x^2+x-2))
What is the first derivative of log_{2}(7x^2+x-2) ?
The first derivative of log_{2}(7x^2+x-2) is (14x+1)/(ln(2)(7x^2+x-2))