What is the derivative of y=3log_{7}(x^2+1) ?

The derivative of y=3log_{7}(x^2+1) is (6x)/(ln(7)(x^2+1))

What is the first derivative of y=3log_{7}(x^2+1) ?

The first derivative of y=3log_{7}(x^2+1) is (6x)/(ln(7)(x^2+1))

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derivative of

y = 3 lo g_{7} (x^{2} + 1)

\frac{6 x}{ln ( 7 ) ( x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (3 lo g_{7} (x^{2} + 1))

Take the constant out:

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

= 3 \frac{d}{d x} (lo g_{7} (x^{2} + 1))

Apply log rule:

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

= 3 \frac{d}{d x} (\frac{ln ( x ^{2} + 1 )}{ln ( 7 )})

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

3 \cdot \frac{1}{ln ( 7 )} \cdot \frac{1}{x ^{2} + 1} \cdot 2 x : \frac{6 x}{ln ( 7 ) ( x ^{2} + 1 )}

= \frac{6 x}{ln ( 7 ) ( x ^{2} + 1 )}

\frac{d y}{d x}, 7 x^{2} - 280 x + y^{2} + 174 = 0

\frac{d}{d x} (sin (x) e^{c o s (x)})

3 t^{4}

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

y = \frac{1}{x ^{2}}, (2, \frac{1}{4})

What is the derivative of y=3log_{7}(x^2+1) ?
The derivative of y=3log_{7}(x^2+1) is (6x)/(ln(7)(x^2+1))
What is the first derivative of y=3log_{7}(x^2+1) ?
The first derivative of y=3log_{7}(x^2+1) is (6x)/(ln(7)(x^2+1))