What is (\partial)/(\partial x)(2yzln(1+x^2)) ?

The answer to (\partial)/(\partial x)(2yzln(1+x^2)) is (4yzx)/(1+x^2)

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(\partial)/(\partial x)(2yzln(1+x^2))

\frac{\partial}{\partial x} (2 yz ln (1 + x^{2}))

\frac{4 yz x}{1 + x ^{2}}

Solution steps

\frac{\partial}{\partial x} (2 yz ln (1 + x^{2}))

Treat

y, z

as constants

Take the constant out:

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

= 2 yz \frac{\partial}{\partial x} (ln (1 + x^{2}))

Apply the chain rule:

\frac{1}{1 + x ^{2}} \frac{\partial}{\partial x} (1 + x^{2})

= \frac{1}{1 + x ^{2}} \frac{\partial}{\partial x} (1 + x^{2})

\frac{\partial}{\partial x} (1 + x^{2}) = 2 x

= 2 yz \frac{1}{1 + x ^{2}} \cdot 2 x

Simplify

2 yz \frac{1}{1 + x ^{2}} \cdot 2 x : \frac{4 yz x}{1 + x ^{2}}

= \frac{4 yz x}{1 + x ^{2}}

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What is (\partial)/(\partial x)(2yzln(1+x^2)) ?
The answer to (\partial)/(\partial x)(2yzln(1+x^2)) is (4yzx)/(1+x^2)