What is (\partial)/(\partial x)((z)(tx^2+yz-2)) ?

The answer to (\partial)/(\partial x)((z)(tx^2+yz-2)) is 2ztx

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(\partial)/(\partial x)((z)(tx^2+yz-2))

\frac{\partial}{\partial x} ((z) (t x^{2} + yz - 2))

2 z t x

Solution steps

\frac{\partial}{\partial x} ((z) (t x^{2} + yz - 2))

Treat

z, t, y

as constants

Take the constant out:

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

= z \frac{\partial}{\partial x} (t x^{2} + yz - 2)

Apply the Sum/Difference Rule:

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

= z (\frac{\partial}{\partial x} (t x^{2}) + \frac{\partial}{\partial x} (yz) - \frac{\partial}{\partial x} (2))

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

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

\frac{\partial}{\partial x} (2) = 0

= z (2 t x + 0 - 0)

Simplify

= 2 z t x

24 (1 - e^{- \frac{t}{2}})

x + 1 \cdot (x + 1)^{2}

(- cos (x) + \frac{cos ^{3} ( x )}{3})^{2}

\frac{sec ^{2} ( \frac{π}{4} )}{2 tan ( \frac{π}{4} )}

x + 1 \cdot (x + 1)

What is (\partial)/(\partial x)((z)(tx^2+yz-2)) ?
The answer to (\partial)/(\partial x)((z)(tx^2+yz-2)) is 2ztx