What is the derivative of 9klog_{e}(((x+9)/(x-9))) ?

The derivative of 9klog_{e}(((x+9)/(x-9))) is -(162k)/((x+9)(x-9))

What is the first derivative of 9klog_{e}(((x+9)/(x-9))) ?

The first derivative of 9klog_{e}(((x+9)/(x-9))) is -(162k)/((x+9)(x-9))

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\frac{d}{d x} (9 k lo g_{e} (\frac{( x + 9 )}{x - 9}))

- \frac{162 k}{( x + 9 ) ( x - 9 )}

Solution steps

\frac{d}{d x} (9 k lo g_{e} (\frac{( x + 9 )}{x - 9}))

Simplify

9 k lo g_{e} (\frac{( x + 9 )}{x - 9}) : 9 k ln (\frac{x + 9}{x - 9})

= \frac{d}{d x} (9 k ln (\frac{x + 9}{x - 9}))

Take the constant out:

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

= 9 k \frac{d}{d x} (ln (\frac{x + 9}{x - 9}))

Apply the chain rule:

\frac{1}{\frac{x + 9}{x - 9}} \frac{d}{d x} (\frac{x + 9}{x - 9})

= \frac{1}{\frac{x + 9}{x - 9}} \frac{d}{d x} (\frac{x + 9}{x - 9})

\frac{d}{d x} (\frac{x + 9}{x - 9}) = - \frac{18}{( x - 9 ) ^{2}}

= 9 k \frac{1}{\frac{x + 9}{x - 9}} (- \frac{18}{( x - 9 ) ^{2}})

Simplify

9 k \frac{1}{\frac{x + 9}{x - 9}} (- \frac{18}{( x - 9 ) ^{2}}) : - \frac{162 k}{( x + 9 ) ( x - 9 )}

= - \frac{162 k}{( x + 9 ) ( x - 9 )}

\frac{5 s + 50.1}{( s + 5 ) ^{2} - 5}

k = 0 \sum \infty \frac{k - 1}{2 ^{k + 1}}

(x, y) \to (1, 2) lim (1 + x + y)

f (x) = x^{3} - 5 + 2 x

x \to \infty lim (e^{- x})

What is the derivative of 9klog_{e}(((x+9)/(x-9))) ?
The derivative of 9klog_{e}(((x+9)/(x-9))) is -(162k)/((x+9)(x-9))
What is the first derivative of 9klog_{e}(((x+9)/(x-9))) ?
The first derivative of 9klog_{e}(((x+9)/(x-9))) is -(162k)/((x+9)(x-9))