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

The derivative of log_{10}(1+x^2y^2) is (2x^2y\frac{dy)/(dx)+2xy^2}{ln(10)(1+x^2y^2)}

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

The first derivative of log_{10}(1+x^2y^2) is (2x^2y\frac{dy)/(dx)+2xy^2}{ln(10)(1+x^2y^2)}

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

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

Solution steps

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

Apply log rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d t} (t^{3} - 6 t^{2} + 9 t)

\frac{d}{d t} (15 + 20 t)

\frac{( 1 + e ^{x} )}{( 1 - e ^{x} )}

y = 3 e^{x} - 14 x - 13

\frac{d}{d s} (x^{2} - 2^{s} in x)

What is the derivative of log_{10}(1+x^2y^2) ?
The derivative of log_{10}(1+x^2y^2) is (2x^2y\frac{dy)/(dx)+2xy^2}{ln(10)(1+x^2y^2)}
What is the first derivative of log_{10}(1+x^2y^2) ?
The first derivative of log_{10}(1+x^2y^2) is (2x^2y\frac{dy)/(dx)+2xy^2}{ln(10)(1+x^2y^2)}