What is (\partial)/(\partial x)(y^{1-x}ln^2(z)) ?

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

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(\partial)/(\partial x)(y^{1-x}ln^2(z))

\frac{\partial}{\partial x} (y^{1 - x} ln^{2} (z))

- y^{1 - x} ln (y) ln^{2} (z)

Solution steps

\frac{\partial}{\partial x} (y^{1 - x} ln^{2} (z))

Treat

y, z

as constants

Take the constant out:

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

= ln^{2} (z) \frac{\partial}{\partial x} (y^{1 - x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

y^{1 - x} = e^{(1 - x) l n (y)}

= ln^{2} (z) \frac{\partial}{\partial x} (e^{(1 - x) l n (y)})

Apply the chain rule:

e^{(1 - x) l n (y)} \frac{\partial}{\partial x} ((1 - x) ln (y))

= e^{(1 - x) l n (y)} \frac{\partial}{\partial x} ((1 - x) ln (y))

\frac{\partial}{\partial x} ((1 - x) ln (y)) = - ln (y)

= ln^{2} (z) e^{(1 - x) l n (y)} (- ln (y))

Simplify

ln^{2} (z) e^{(1 - x) l n (y)} (- ln (y)) : - y^{1 - x} ln (y) ln^{2} (z)

= - y^{1 - x} ln (y) ln^{2} (z)

\frac{d}{d x} (c \cdot ln (x))

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

x^{3} - 4 x, 12 x

f (x) = x^{2} - 4, at x = 2

f (x) = \frac{6 x}{x ^{2} + 1}, at x = 0

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