What is (\partial)/(\partial t)(x^2ycos(z/t)) ?

The answer to (\partial)/(\partial t)(x^2ycos(z/t)) is (x^2yzsin(z/t))/(t^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial t)(x^2ycos(z/t))

\frac{\partial}{\partial t} (x^{2} y cos (\frac{z}{t}))

\frac{x ^{2} yz sin ( \frac{z}{t} )}{t ^{2}}

Solution steps

\frac{\partial}{\partial t} (x^{2} y cos (\frac{z}{t}))

Treat

x, y, z

as constants

Take the constant out:

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

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

Apply the chain rule:

- sin (\frac{z}{t}) \frac{\partial}{\partial t} (\frac{z}{t})

= - sin (\frac{z}{t}) \frac{\partial}{\partial t} (\frac{z}{t})

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

= x^{2} y (- sin (\frac{z}{t}) (- \frac{z}{t ^{2}}))

Simplify

x^{2} y (- sin (\frac{z}{t}) (- \frac{z}{t ^{2}})) : \frac{x ^{2} yz sin ( \frac{z}{t} )}{t ^{2}}

= \frac{x ^{2} yz sin ( \frac{z}{t} )}{t ^{2}}

\int sec^{5} (x) tan (x) d x

f (x) = - \frac{13}{x ^{6}}

\int \frac{3}{t t ^{2} - 1} d t

e^{- x^{2}}

x \to + 0 lim (\frac{x}{x - 1})

What is (\partial)/(\partial t)(x^2ycos(z/t)) ?
The answer to (\partial)/(\partial t)(x^2ycos(z/t)) is (x^2yzsin(z/t))/(t^2)