What is the derivative of y=sqrt(-2x^2+32x-7) ?

The derivative of y=sqrt(-2x^2+32x-7) is (2(-x+8))/(sqrt(-2x^2+32x-7))

What is the first derivative of y=sqrt(-2x^2+32x-7) ?

The first derivative of y=sqrt(-2x^2+32x-7) is (2(-x+8))/(sqrt(-2x^2+32x-7))

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derivative of

y = - 2 x^{2} + 32 x - 7

\frac{2 ( - x + 8 )}{- 2 x ^{2} + 32 x - 7}

Solution steps

\frac{d}{d x} (- 2 x^{2} + 32 x - 7)

Apply the chain rule:

\frac{1}{2 - 2 x ^{2} + 32 x - 7} \frac{d}{d x} (- 2 x^{2} + 32 x - 7)

= \frac{1}{2 - 2 x ^{2} + 32 x - 7} \frac{d}{d x} (- 2 x^{2} + 32 x - 7)

\frac{d}{d x} (- 2 x^{2} + 32 x - 7) = - 4 x + 32

= \frac{1}{2 - 2 x ^{2} + 32 x - 7} (- 4 x + 32)

Simplify

\frac{1}{2 - 2 x ^{2} + 32 x - 7} (- 4 x + 32) : \frac{2 ( - x + 8 )}{- 2 x ^{2} + 32 x - 7}

= \frac{2 ( - x + 8 )}{- 2 x ^{2} + 32 x - 7}

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

\int 2^{t} d t

\frac{1}{x ^{2} - x + 1}

\frac{\partial}{\partial x} (x y^{2} z^{3} + 3 yz)

\frac{\partial}{\partial y} (x^{3} + y^{2} - 5 x y)

What is the derivative of y=sqrt(-2x^2+32x-7) ?
The derivative of y=sqrt(-2x^2+32x-7) is (2(-x+8))/(sqrt(-2x^2+32x-7))
What is the first derivative of y=sqrt(-2x^2+32x-7) ?
The first derivative of y=sqrt(-2x^2+32x-7) is (2(-x+8))/(sqrt(-2x^2+32x-7))