What is the derivative of y=sqrt(1+3x^2)(ln(x^2)) ?

The derivative of y=sqrt(1+3x^2)(ln(x^2)) is (3xln(x^2)}{sqrt(1+3x^2)}+\frac{2sqrt(1+3x^2))/x

What is the first derivative of y=sqrt(1+3x^2)(ln(x^2)) ?

The first derivative of y=sqrt(1+3x^2)(ln(x^2)) is (3xln(x^2)}{sqrt(1+3x^2)}+\frac{2sqrt(1+3x^2))/x

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derivative of

y = 1 + 3 x^{2} (ln (x^{2}))

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

Solution steps

\frac{d}{d x} (1 + 3 x^{2} ln (x^{2}))

Apply the Product Rule:

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

f = 1 + 3 x^{2}, g = ln (x^{2})

= \frac{d}{d x} (1 + 3 x^{2}) ln (x^{2}) + \frac{d}{d x} (ln (x^{2})) 1 + 3 x^{2}

\frac{d}{d x} (1 + 3 x^{2}) = \frac{3 x}{1 + 3 x ^{2}}

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

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

Simplify

\frac{3 x}{1 + 3 x ^{2}} ln (x^{2}) + \frac{2}{x} 1 + 3 x^{2} : \frac{3 x ln ( x ^{2} )}{1 + 3 x ^{2}} + \frac{2 1 + 3 x ^{2}}{x}

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

f (x) = (x^{3} + 4) e^{x}

\int \frac{x ^{2} + 4}{x + 1} d x

n = 0 \sum \infty 3 (\frac{x}{4})^{n}

f (x) = \frac{1}{x - 5}, at x = 10

g (x) = x^{2} (1 - 7 x)

What is the derivative of y=sqrt(1+3x^2)(ln(x^2)) ?
The derivative of y=sqrt(1+3x^2)(ln(x^2)) is (3xln(x^2)}{sqrt(1+3x^2)}+\frac{2sqrt(1+3x^2))/x
What is the first derivative of y=sqrt(1+3x^2)(ln(x^2)) ?
The first derivative of y=sqrt(1+3x^2)(ln(x^2)) is (3xln(x^2)}{sqrt(1+3x^2)}+\frac{2sqrt(1+3x^2))/x