What is (\partial)/(\partial x)((x-1)^2+8(y-1)^2) ?

The answer to (\partial)/(\partial x)((x-1)^2+8(y-1)^2) is 2x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)((x-1)^2+8(y-1)^2)

\frac{\partial}{\partial x} ((x - 1)^{2} + 8 (y - 1)^{2})

2 x - 2

Solution steps

\frac{\partial}{\partial x} ((x - 1)^{2} + 8 (y - 1)^{2})

Simplify

(x - 1)^{2} + 8 (y - 1)^{2} : x^{2} - 2 x + 8 y^{2} + 9 - 16 y

= \frac{\partial}{\partial x} (x^{2} - 2 x + 8 y^{2} + 9 - 16 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (2 x) + \frac{\partial}{\partial x} (8 y^{2}) + \frac{\partial}{\partial x} (9) - \frac{\partial}{\partial x} (16 y)

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (2 x) = 2

\frac{\partial}{\partial x} (8 y^{2}) = 0

\frac{\partial}{\partial x} (9) = 0

\frac{\partial}{\partial x} (16 y) = 0

= 2 x - 2 + 0 + 0 - 0

Simplify

= 2 x - 2

x \to \frac{π}{4} lim ((tan (x))^{t a n (2 x)})

x \to 5 + lim (ln (x^{2} - 25))

y = x 16 - x^{2}

\frac{d}{d x} (ln (cos^{2} (x)))

y^{^{''}} + y = 3 x^{2}

What is (\partial)/(\partial x)((x-1)^2+8(y-1)^2) ?
The answer to (\partial)/(\partial x)((x-1)^2+8(y-1)^2) is 2x-2