What is the derivative of log_{10}(2x^2+5x+2) ?

The derivative of log_{10}(2x^2+5x+2) is (4x+5)/(ln(10)(2x^2+5x+2))

What is the first derivative of log_{10}(2x^2+5x+2) ?

The first derivative of log_{10}(2x^2+5x+2) is (4x+5)/(ln(10)(2x^2+5x+2))

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\frac{d}{d x} (lo g_{10} (2 x^{2} + 5 x + 2))

\frac{4 x + 5}{ln ( 10 ) ( 2 x ^{2} + 5 x + 2 )}

Solution steps

\frac{d}{d x} (lo g_{10} (2 x^{2} + 5 x + 2))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 2 x ^{2} + 5 x + 2 )}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (2 x^{2} + 5 x + 2))

Apply the chain rule:

\frac{1}{2 x ^{2} + 5 x + 2} \frac{d}{d x} (2 x^{2} + 5 x + 2)

= \frac{1}{2 x ^{2} + 5 x + 2} \frac{d}{d x} (2 x^{2} + 5 x + 2)

\frac{d}{d x} (2 x^{2} + 5 x + 2) = 4 x + 5

= \frac{1}{ln ( 10 )} \cdot \frac{1}{2 x ^{2} + 5 x + 2} (4 x + 5)

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{2 x ^{2} + 5 x + 2} (4 x + 5) : \frac{4 x + 5}{ln ( 10 ) ( 2 x ^{2} + 5 x + 2 )}

= \frac{4 x + 5}{ln ( 10 ) ( 2 x ^{2} + 5 x + 2 )}

\frac{3 x - y}{x}

\frac{\partial}{\partial x} (x^{0.2} y^{0.4})

x \to 3 lim (\frac{3 - x}{x ^{2} - 9})

\int \frac{1}{x ^{2} + 4 x + 20} d x

f (t) = t sin (π t)

What is the derivative of log_{10}(2x^2+5x+2) ?
The derivative of log_{10}(2x^2+5x+2) is (4x+5)/(ln(10)(2x^2+5x+2))
What is the first derivative of log_{10}(2x^2+5x+2) ?
The first derivative of log_{10}(2x^2+5x+2) is (4x+5)/(ln(10)(2x^2+5x+2))