What is (\partial)/(\partial x)(-tan(x^3+y)) ?

The answer to (\partial)/(\partial x)(-tan(x^3+y)) is -sec^2(x^3+y)*3x^2

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(\partial)/(\partial x)(-tan(x^3+y))

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

- sec^{2} (x^{3} + y) \cdot 3 x^{2}

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{\partial}{\partial x} (tan (x^{3} + y))

Apply the chain rule:

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

= sec^{2} (x^{3} + y) \frac{\partial}{\partial x} (x^{3} + y)

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

= - sec^{2} (x^{3} + y) \cdot 3 x^{2}

x \to 0 lim (\frac{2 - ∣ x ∣}{2 + x})

n = 0 \sum \infty n sin (π \cdot n)

ln (7)

\frac{d}{d x} (sin (12 x))

\int_{2}^{8} t^{3} ln (2 t) d t

What is (\partial)/(\partial x)(-tan(x^3+y)) ?
The answer to (\partial)/(\partial x)(-tan(x^3+y)) is -sec^2(x^3+y)*3x^2