What is (\partial)/(\partial z)(ln(x^2+y^4)-z) ?

The answer to (\partial)/(\partial z)(ln(x^2+y^4)-z) is -1

Popular >

(\partial)/(\partial z)(ln(x^2+y^4)-z)

\frac{\partial}{\partial z} (ln (x^{2} + y^{4}) - z)

- 1

Solution steps

\frac{\partial}{\partial z} (ln (x^{2} + y^{4}) - z)

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial z} (ln (x^{2} + y^{4})) - \frac{\partial}{\partial z} (z)

\frac{\partial}{\partial z} (ln (x^{2} + y^{4})) = 0

\frac{\partial}{\partial z} (z) = 1

= 0 - 1

Simplify

= - 1

25 y^{^{''}} - y = x e^{\frac{x}{5}}, y (0) = 1, y^{^{'}} (0) = 0

x \to \infty lim (\frac{1}{x ( ln ( x ) ) ^{4}})

\int_{1}^{3} 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

\frac{\partial}{\partial v} (uv)

\frac{\partial}{\partial z} (2 x^{4} y z^{3})

What is (\partial)/(\partial z)(ln(x^2+y^4)-z) ?
The answer to (\partial)/(\partial z)(ln(x^2+y^4)-z) is -1