What is the derivative of y=log_{5}(sqrt(7x+1)) ?

The derivative of y=log_{5}(sqrt(7x+1)) is 7/(2ln(5)(7x+1))

What is the first derivative of y=log_{5}(sqrt(7x+1)) ?

The first derivative of y=log_{5}(sqrt(7x+1)) is 7/(2ln(5)(7x+1))

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derivative of

y = lo g_{5} (7 x + 1)

\frac{7}{2 ln ( 5 ) ( 7 x + 1 )}

Solution steps

\frac{d}{d x} (lo g_{5} (7 x + 1))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 7 x + 1 )}{ln ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (7 x + 1))

Apply the chain rule:

\frac{1}{7 x + 1} \frac{d}{d x} (7 x + 1)

= \frac{1}{7 x + 1} \frac{d}{d x} (7 x + 1)

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

= \frac{1}{ln ( 5 )} \cdot \frac{1}{7 x + 1} \cdot \frac{7}{2 7 x + 1}

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{7 x + 1} \cdot \frac{7}{2 7 x + 1} : \frac{7}{2 ln ( 5 ) ( 7 x + 1 )}

= \frac{7}{2 ln ( 5 ) ( 7 x + 1 )}

\frac{d}{d x} (\frac{4 x ^{3}}{3})

\frac{\partial}{\partial y} (z cos (x z))

\frac{d}{d x} ((x - 2)^{4} - 1)

\int x I n (1 + x)

(- 6, 7), (1, - 12)

What is the derivative of y=log_{5}(sqrt(7x+1)) ?
The derivative of y=log_{5}(sqrt(7x+1)) is 7/(2ln(5)(7x+1))
What is the first derivative of y=log_{5}(sqrt(7x+1)) ?
The first derivative of y=log_{5}(sqrt(7x+1)) is 7/(2ln(5)(7x+1))