What is the derivative of e^{6x}ln(x^2-4) ?

The derivative of e^{6x}ln(x^2-4) is 6e^{6x}ln(x^2-4)+(2e^{6x}x)/(x^2-4)

What is the first derivative of e^{6x}ln(x^2-4) ?

The first derivative of e^{6x}ln(x^2-4) is 6e^{6x}ln(x^2-4)+(2e^{6x}x)/(x^2-4)

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\frac{d}{d x} (e^{6 x} ln (x^{2} - 4))

6 e^{6 x} ln (x^{2} - 4) + \frac{2 e ^{6 x} x}{x ^{2} - 4}

Solution steps

\frac{d}{d x} (e^{6 x} ln (x^{2} - 4))

Apply the Product Rule:

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

f = e^{6 x}, g = ln (x^{2} - 4)

= \frac{d}{d x} (e^{6 x}) ln (x^{2} - 4) + \frac{d}{d x} (ln (x^{2} - 4)) e^{6 x}

\frac{d}{d x} (e^{6 x}) = e^{6 x} \cdot 6

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

= e^{6 x} \cdot 6 ln (x^{2} - 4) + \frac{2 x}{x ^{2} - 4} e^{6 x}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 6 e^{6 x} ln (x^{2} - 4) + \frac{2 e ^{6 x} x}{x ^{2} - 4}

\int_{- 1}^{2} (x - 6 ∣ x ∣) d x

\int \frac{1}{- x ^{2} + 6 x - 5} d x

4 x^{3} y + x^{4} y^{^{'}} = sin^{3} (\frac{x}{9})

ln (x + 1), 0

\frac{d}{d x} (3 sin (2 x) cos (x))

What is the derivative of e^{6x}ln(x^2-4) ?
The derivative of e^{6x}ln(x^2-4) is 6e^{6x}ln(x^2-4)+(2e^{6x}x)/(x^2-4)
What is the first derivative of e^{6x}ln(x^2-4) ?
The first derivative of e^{6x}ln(x^2-4) is 6e^{6x}ln(x^2-4)+(2e^{6x}x)/(x^2-4)