What is the derivative of log_{4}(x^3-7x+5) ?

The derivative of log_{4}(x^3-7x+5) is (3x^2-7)/(2ln(2)(x^3-7x+5))

What is the first derivative of log_{4}(x^3-7x+5) ?

The first derivative of log_{4}(x^3-7x+5) is (3x^2-7)/(2ln(2)(x^3-7x+5))

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\frac{d}{d x} (lo g_{4} (x^{3} - 7 x + 5))

\frac{3 x ^{2} - 7}{2 ln ( 2 ) ( x ^{3} - 7 x + 5 )}

Solution steps

\frac{d}{d x} (lo g_{4} (x^{3} - 7 x + 5))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{3} - 7 x + 5 )}{ln ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (x^{3} - 7 x + 5))

Apply the chain rule:

\frac{1}{x ^{3} - 7 x + 5} \frac{d}{d x} (x^{3} - 7 x + 5)

= \frac{1}{x ^{3} - 7 x + 5} \frac{d}{d x} (x^{3} - 7 x + 5)

\frac{d}{d x} (x^{3} - 7 x + 5) = 3 x^{2} - 7

= \frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{3} - 7 x + 5} (3 x^{2} - 7)

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{3} - 7 x + 5} (3 x^{2} - 7) : \frac{3 x ^{2} - 7}{2 ln ( 2 ) ( x ^{3} - 7 x + 5 )}

= \frac{3 x ^{2} - 7}{2 ln ( 2 ) ( x ^{3} - 7 x + 5 )}

\int_{0}^{π} - sin (x) d x

x \to 0 lim ((sin (x))^{t a n (x)})

f (x) = sin (sin (x))

\frac{d}{d x} (\frac{1}{4} x - \frac{1}{6})

\int \frac{8 x ^{3} - 16 x ^{2} + 5}{x ^{2} - 2 x} d x

What is the derivative of log_{4}(x^3-7x+5) ?
The derivative of log_{4}(x^3-7x+5) is (3x^2-7)/(2ln(2)(x^3-7x+5))
What is the first derivative of log_{4}(x^3-7x+5) ?
The first derivative of log_{4}(x^3-7x+5) is (3x^2-7)/(2ln(2)(x^3-7x+5))