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

The derivative of y=(4x+1)^2 is 8(4x+1)

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

The first derivative of y=(4x+1)^2 is 8(4x+1)

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derivative of

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

8 (4 x + 1)

Solution steps

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

Apply the chain rule:

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

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

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

= 2 (4 x + 1) \cdot 4

Simplify

= 8 (4 x + 1)

x \to 0 lim (\frac{sin ( x ) - 1}{x})

x \to 0 lim (\frac{1}{x} - \frac{1}{x ^{2}} - x)

y^{^{''}} - (\frac{6}{x}) y^{^{'}} - (\frac{3}{x ^{2}}) y = 0

n = 1 \sum \infty n - 1

\frac{d}{d x} ((sin (x))^{6})

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