What is the derivative of g(x)=log_{7}(8x^3-9x^2) ?

The derivative of g(x)=log_{7}(8x^3-9x^2) is (6(4x-3))/(ln(7)x(8x-9))

What is the first derivative of g(x)=log_{7}(8x^3-9x^2) ?

The first derivative of g(x)=log_{7}(8x^3-9x^2) is (6(4x-3))/(ln(7)x(8x-9))

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derivative of

g (x) = lo g_{7} (8 x^{3} - 9 x^{2})

\frac{6 ( 4 x - 3 )}{ln ( 7 ) x ( 8 x - 9 )}

Solution steps

\frac{d}{d x} (lo g_{7} (8 x^{3} - 9 x^{2}))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 8 x ^{3} - 9 x ^{2} )}{ln ( 7 )})

Take the constant out:

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

= \frac{1}{ln ( 7 )} \frac{d}{d x} (ln (8 x^{3} - 9 x^{2}))

Apply the chain rule:

\frac{1}{8 x ^{3} - 9 x ^{2}} \frac{d}{d x} (8 x^{3} - 9 x^{2})

= \frac{1}{8 x ^{3} - 9 x ^{2}} \frac{d}{d x} (8 x^{3} - 9 x^{2})

\frac{d}{d x} (8 x^{3} - 9 x^{2}) = 24 x^{2} - 18 x

= \frac{1}{ln ( 7 )} \cdot \frac{1}{8 x ^{3} - 9 x ^{2}} (24 x^{2} - 18 x)

Simplify

\frac{1}{ln ( 7 )} \cdot \frac{1}{8 x ^{3} - 9 x ^{2}} (24 x^{2} - 18 x) : \frac{6 ( 4 x - 3 )}{ln ( 7 ) x ( 8 x - 9 )}

= \frac{6 ( 4 x - 3 )}{ln ( 7 ) x ( 8 x - 9 )}

\int x (x sin (x) - cos (x)) d x

\int_{0}^{0.9} \frac{1}{( 1 - x )} d x

\int 6 cos (5 x) d x

y = 2 (x + y) y^{^{'}}

\int 3 sin^{6} (x) cos (x) d x

What is the derivative of g(x)=log_{7}(8x^3-9x^2) ?
The derivative of g(x)=log_{7}(8x^3-9x^2) is (6(4x-3))/(ln(7)x(8x-9))
What is the first derivative of g(x)=log_{7}(8x^3-9x^2) ?
The first derivative of g(x)=log_{7}(8x^3-9x^2) is (6(4x-3))/(ln(7)x(8x-9))