What is the derivative of y=(3x+1)sqrt(8x-1) ?

The derivative of y=(3x+1)sqrt(8x-1) is (36x+1)/(sqrt(8x-1))

What is the first derivative of y=(3x+1)sqrt(8x-1) ?

The first derivative of y=(3x+1)sqrt(8x-1) is (36x+1)/(sqrt(8x-1))

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derivative of

y = (3 x + 1) 8 x - 1

\frac{36 x + 1}{8 x - 1}

Solution steps

\frac{d}{d x} ((3 x + 1) 8 x - 1)

Apply the Product Rule:

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

f = 3 x + 1, g = 8 x - 1

= \frac{d}{d x} (3 x + 1) 8 x - 1 + \frac{d}{d x} (8 x - 1) (3 x + 1)

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

\frac{d}{d x} (8 x - 1) = \frac{4}{8 x - 1}

= 3 8 x - 1 + \frac{4}{8 x - 1} (3 x + 1)

Simplify

3 8 x - 1 + \frac{4}{8 x - 1} (3 x + 1) : \frac{36 x + 1}{8 x - 1}

= \frac{36 x + 1}{8 x - 1}

\frac{d y}{d x} = x y

x \to \infty lim (5 x + 2)

\frac{d}{d x} y = \frac{1}{( 8 x - 5 ) ^{5}}

y^{^{'}} - 2 y = 4 x

\int \frac{( 1 + cos ( x ) )}{x + sin ( x )} d x

What is the derivative of y=(3x+1)sqrt(8x-1) ?
The derivative of y=(3x+1)sqrt(8x-1) is (36x+1)/(sqrt(8x-1))
What is the first derivative of y=(3x+1)sqrt(8x-1) ?
The first derivative of y=(3x+1)sqrt(8x-1) is (36x+1)/(sqrt(8x-1))