What is the derivative of 5xlog_{4}(2x-1) ?

The derivative of 5xlog_{4}(2x-1) is 5(log_{4}(2x-1)+x/(ln(2)(2x-1)))

What is the first derivative of 5xlog_{4}(2x-1) ?

The first derivative of 5xlog_{4}(2x-1) is 5(log_{4}(2x-1)+x/(ln(2)(2x-1)))

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\frac{d}{d x} (5 x lo g_{4} (2 x - 1))

5 (lo g_{4} (2 x - 1) + \frac{x}{ln ( 2 ) ( 2 x - 1 )})

Solution steps

\frac{d}{d x} (5 x lo g_{4} (2 x - 1))

Take the constant out:

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

= 5 \frac{d}{d x} (x lo g_{4} (2 x - 1))

Apply the Product Rule:

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

f = x, g = lo g_{4} (2 x - 1)

= 5 (\frac{d x}{d x} lo g_{4} (2 x - 1) + \frac{d}{d x} (lo g_{4} (2 x - 1)) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (lo g_{4} (2 x - 1)) = \frac{1}{ln ( 2 ) ( 2 x - 1 )}

= 5 (1 \cdot lo g_{4} (2 x - 1) + \frac{1}{ln ( 2 ) ( 2 x - 1 )} x)

Simplify

5 (1 \cdot lo g_{4} (2 x - 1) + \frac{1}{ln ( 2 ) ( 2 x - 1 )} x) : 5 (lo g_{4} (2 x - 1) + \frac{x}{ln ( 2 ) ( 2 x - 1 )})

= 5 (lo g_{4} (2 x - 1) + \frac{x}{ln ( 2 ) ( 2 x - 1 )})

\frac{\partial}{\partial y} ((x - y)^{2})

a

y^{^{''}} - 8 y^{^{'}} + 14 y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{\partial ^{2}}{\partial x ^{2}} (\frac{x y}{x - y})

y^{^{''}} - 16 y^{^{'}} + 64 y = 5 t^{- 1} e^{8 t}

What is the derivative of 5xlog_{4}(2x-1) ?
The derivative of 5xlog_{4}(2x-1) is 5(log_{4}(2x-1)+x/(ln(2)(2x-1)))
What is the first derivative of 5xlog_{4}(2x-1) ?
The first derivative of 5xlog_{4}(2x-1) is 5(log_{4}(2x-1)+x/(ln(2)(2x-1)))