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

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

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

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

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derivative of

lo g_{10} (x^{2} + y^{2})

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

Solution steps

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

Treat

y

as a constant

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

3^{x + 2}

3^{x + 1}

4 x ∣ x ∣

3 x - cos (x) - 1

\frac{d}{d t} (10 + 2 t)

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