What is the derivative of f(x)=ln(4x^2+1) ?

The derivative of f(x)=ln(4x^2+1) is (8x)/(4x^2+1)

What is the first derivative of f(x)=ln(4x^2+1) ?

The first derivative of f(x)=ln(4x^2+1) is (8x)/(4x^2+1)

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derivative of

f (x) = ln (4 x^{2} + 1)

\frac{8 x}{4 x ^{2} + 1}

Solution steps

\frac{d}{d x} (ln (4 x^{2} + 1))

Apply the chain rule:

\frac{1}{4 x ^{2} + 1} \frac{d}{d x} (4 x^{2} + 1)

= \frac{1}{4 x ^{2} + 1} \frac{d}{d x} (4 x^{2} + 1)

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= \frac{1}{4 x ^{2} + 1} \cdot 8 x

Simplify

\frac{1}{4 x ^{2} + 1} \cdot 8 x : \frac{8 x}{4 x ^{2} + 1}

= \frac{8 x}{4 x ^{2} + 1}

x \to 1 - lim (\frac{2 x}{x ^{2} - 1})

\frac{\partial}{\partial y} (x^{2} y + x + 1)

2^{30}

\int sin (\frac{x}{4}) d x

\int \frac{x ^{3}}{( x ^{4} + 3 ) ^{(\frac{7}{2})}} d x

What is the derivative of f(x)=ln(4x^2+1) ?
The derivative of f(x)=ln(4x^2+1) is (8x)/(4x^2+1)
What is the first derivative of f(x)=ln(4x^2+1) ?
The first derivative of f(x)=ln(4x^2+1) is (8x)/(4x^2+1)