What is the derivative of log_{10}((x^2-1/x)) ?

The derivative of log_{10}((x^2-1/x)) is (x^2+1)/(ln(10)x(x^2-1))

What is the first derivative of log_{10}((x^2-1/x)) ?

The first derivative of log_{10}((x^2-1/x)) is (x^2+1)/(ln(10)x(x^2-1))

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\frac{d}{d x} (lo g_{10} (\frac{x ^{2} - 1}{x}))

\frac{x ^{2} + 1}{ln ( 10 ) x ( x ^{2} - 1 )}

Solution steps

\frac{d}{d x} (lo g_{10} (\frac{x ^{2} - 1}{x}))

Apply log rule:

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

= \frac{d}{d x} \frac{ln ( \frac{x ^{2} - 1}{x} )}{ln ( 10 )}

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (\frac{x ^{2} - 1}{x}))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{\frac{x ^{2} - 1}{x}} \cdot \frac{x ^{2} + 1}{x ^{2}}

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{\frac{x ^{2} - 1}{x}} \cdot \frac{x ^{2} + 1}{x ^{2}} : \frac{x ^{2} + 1}{ln ( 10 ) x ( x ^{2} - 1 )}

= \frac{x ^{2} + 1}{ln ( 10 ) x ( x ^{2} - 1 )}

\int (ln (x)) (sin (x)) d x

\frac{d r}{d θ} cot (θ) - r = 2

f (x) = x^{2} - 8, (- 4, 8)

\int_{1}^{4} - 8 t ln (t) d t

\int \frac{e ^{- 2 x}}{1 + e ^{- 2 x}} d x

What is the derivative of log_{10}((x^2-1/x)) ?
The derivative of log_{10}((x^2-1/x)) is (x^2+1)/(ln(10)x(x^2-1))
What is the first derivative of log_{10}((x^2-1/x)) ?
The first derivative of log_{10}((x^2-1/x)) is (x^2+1)/(ln(10)x(x^2-1))