What is the derivative of (16cx-4x^2^{1/2}) ?

The derivative of (16cx-4x^2^{1/2}) is (2(-x+2c))/((x(-x+4c))^{1/2)}

What is the first derivative of (16cx-4x^2^{1/2}) ?

The first derivative of (16cx-4x^2^{1/2}) is (2(-x+2c))/((x(-x+4c))^{1/2)}

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\frac{d}{d x} ((16 c x - 4 x^{2})^{\frac{1}{2}})

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

Solution steps

\frac{d}{d x} ((16 c x - 4 x^{2})^{\frac{1}{2}})

Apply the chain rule:

\frac{1}{2 ( 16 c x - 4 x ^{2} ) ^{\frac{1}{2}}} \frac{d}{d x} (16 c x - 4 x^{2})

= \frac{1}{2 ( 16 c x - 4 x ^{2} ) ^{\frac{1}{2}}} \frac{d}{d x} (16 c x - 4 x^{2})

\frac{d}{d x} (16 c x - 4 x^{2}) = 16 c - 8 x

= \frac{1}{2 ( 16 c x - 4 x ^{2} ) ^{\frac{1}{2}}} (16 c - 8 x)

Simplify

\frac{1}{2 ( 16 c x - 4 x ^{2} ) ^{\frac{1}{2}}} (16 c - 8 x) : \frac{2 ( - x + 2 c )}{( x ( - x + 4 c ) ) ^{\frac{1}{2}}}

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

x \to 0 lim (x^{4} - \frac{x ^{3}}{3})

y = 3 x^{2}, y = 18 x - 27, [0, 3]

\int sin^{2} (x) cos (z) d z

\frac{d}{d x} (- 7 x^{2})

f (x) = (3 x - 1) (x + 2)

What is the derivative of (16cx-4x^2^{1/2}) ?
The derivative of (16cx-4x^2^{1/2}) is (2(-x+2c))/((x(-x+4c))^{1/2)}
What is the first derivative of (16cx-4x^2^{1/2}) ?
The first derivative of (16cx-4x^2^{1/2}) is (2(-x+2c))/((x(-x+4c))^{1/2)}