What is (\partial)/(\partial x)((a-x)^2+b(y-x^2)^2) ?

The answer to (\partial)/(\partial x)((a-x)^2+b(y-x^2)^2) is -2(a-x)-4bx(y-x^2)

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(\partial)/(\partial x)((a-x)^2+b(y-x^2)^2)

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

- 2 (a - x) - 4 b x (y - x^{2})

Solution steps

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

Treat

a, b, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} ((a - x)^{2}) + \frac{\partial}{\partial x} (b (y - x^{2})^{2})

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

\frac{\partial}{\partial x} (b (y - x^{2})^{2}) = - 4 b x (y - x^{2})

= - 2 (a - x) - 4 b x (y - x^{2})

\int 6 x^{2} y d y

\frac{\partial}{\partial L} (b L^{3} K (b) (b)^{2})

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

\frac{d}{d x} (6 x^{\frac{3}{2}} sec (x))

What is (\partial)/(\partial x)((a-x)^2+b(y-x^2)^2) ?
The answer to (\partial)/(\partial x)((a-x)^2+b(y-x^2)^2) is -2(a-x)-4bx(y-x^2)