What is the derivative of y=5ln(4x^2+1) ?

The derivative of y=5ln(4x^2+1) is (40x)/(4x^2+1)

What is the first derivative of y=5ln(4x^2+1) ?

The first derivative of y=5ln(4x^2+1) is (40x)/(4x^2+1)

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derivative of

y = 5 ln (4 x^{2} + 1)

\frac{40 x}{4 x ^{2} + 1}

Solution steps

\frac{d}{d x} (5 ln (4 x^{2} + 1))

Take the constant out:

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

= 5 \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

= 5 \cdot \frac{1}{4 x ^{2} + 1} \cdot 8 x

Simplify

5 \cdot \frac{1}{4 x ^{2} + 1} \cdot 8 x : \frac{40 x}{4 x ^{2} + 1}

= \frac{40 x}{4 x ^{2} + 1}

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1 - x

What is the derivative of y=5ln(4x^2+1) ?
The derivative of y=5ln(4x^2+1) is (40x)/(4x^2+1)
What is the first derivative of y=5ln(4x^2+1) ?
The first derivative of y=5ln(4x^2+1) is (40x)/(4x^2+1)