What is the derivative of log_{2}(2x^2-7x+1) ?

The derivative of log_{2}(2x^2-7x+1) is (4x-7)/(ln(2)(2x^2-7x+1))

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

The first derivative of log_{2}(2x^2-7x+1) is (4x-7)/(ln(2)(2x^2-7x+1))

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

\frac{4 x - 7}{ln ( 2 ) ( 2 x ^{2} - 7 x + 1 )}

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (2 x^{2} - 7 x + 1) = 4 x - 7

= \frac{1}{ln ( 2 )} \cdot \frac{1}{2 x ^{2} - 7 x + 1} (4 x - 7)

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{2 x ^{2} - 7 x + 1} (4 x - 7) : \frac{4 x - 7}{ln ( 2 ) ( 2 x ^{2} - 7 x + 1 )}

= \frac{4 x - 7}{ln ( 2 ) ( 2 x ^{2} - 7 x + 1 )}

\int 4 x e^{9 x} d x

f (x) = x^{4} - 5 x^{2} - x, at x = 1

y = \frac{3}{x}, (5, \frac{3}{5})

f (x) = (2 x + 3) (x^{2} + 2 x)

\int_{0}^{5} x e^{- 0.6 x} d x

What is the derivative of log_{2}(2x^2-7x+1) ?
The derivative of log_{2}(2x^2-7x+1) is (4x-7)/(ln(2)(2x^2-7x+1))
What is the first derivative of log_{2}(2x^2-7x+1) ?
The first derivative of log_{2}(2x^2-7x+1) is (4x-7)/(ln(2)(2x^2-7x+1))