What is the derivative of y=(9x)^{3x} ?

The derivative of y=(9x)^{3x} is 2ln(3)*3^{6x+1}x^{3x}+3^{1+6x}x^{3x}(ln(x)+1)

What is the first derivative of y=(9x)^{3x} ?

The first derivative of y=(9x)^{3x} is 2ln(3)*3^{6x+1}x^{3x}+3^{1+6x}x^{3x}(ln(x)+1)

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derivative of

y = (9 x)^{3 x}

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

Solution steps

\frac{d}{d x} ((9 x)^{3 x})

Simplify

(9 x)^{3 x} : 72 9^{x} x^{3 x}

= \frac{d}{d x} (72 9^{x} x^{3 x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 72 9^{x}, g = x^{3 x}

= \frac{d}{d x} (72 9^{x}) x^{3 x} + \frac{d}{d x} (x^{3 x}) 72 9^{x}

\frac{d}{d x} (72 9^{x}) = 2 ln (3) \cdot 3^{6 x + 1}

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

= 2 ln (3) \cdot 3^{6 x + 1} x^{3 x} + 3 x^{3 x} (ln (x) + 1) \cdot 72 9^{x}

3 x^{3 x} (ln (x) + 1) \cdot 72 9^{x} = 3^{1 + 6 x} x^{3 x} (ln (x) + 1)

= 2 ln (3) \cdot 3^{6 x + 1} x^{3 x} + 3^{1 + 6 x} x^{3 x} (ln (x) + 1)

\int \frac{1}{x ( x - 5 ) ^{2}} d x

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

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

\int (2 x - 1) (3 x + 4) d x

\int 0.002 d x

What is the derivative of y=(9x)^{3x} ?
The derivative of y=(9x)^{3x} is 2ln(3)*3^{6x+1}x^{3x}+3^{1+6x}x^{3x}(ln(x)+1)
What is the first derivative of y=(9x)^{3x} ?
The first derivative of y=(9x)^{3x} is 2ln(3)*3^{6x+1}x^{3x}+3^{1+6x}x^{3x}(ln(x)+1)