What is (\partial)/(\partial x)(5ln(1+x^2-y)) ?

The answer to (\partial)/(\partial x)(5ln(1+x^2-y)) is (10x)/(1+x^2-y)

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(\partial)/(\partial x)(5ln(1+x^2-y))

\frac{\partial}{\partial x} (5 ln (1 + x^{2} - y))

\frac{10 x}{1 + x ^{2} - y}

Solution steps

\frac{\partial}{\partial x} (5 ln (1 + x^{2} - y))

Treat

y

as a constant

Take the constant out:

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

= 5 \frac{\partial}{\partial x} (ln (1 + x^{2} - y))

Apply the chain rule:

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

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

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

= 5 \cdot \frac{1}{1 + x ^{2} - y} \cdot 2 x

Simplify

5 \cdot \frac{1}{1 + x ^{2} - y} \cdot 2 x : \frac{10 x}{1 + x ^{2} - y}

= \frac{10 x}{1 + x ^{2} - y}

x \to 0 lim (\frac{ln ( x )}{x ^{2}})

\int \frac{1}{1 + x + x ^{2}} d x

f^{^{'}} (x) = 5 (1 + e^{x})^{4} e^{x}

h (w) = \frac{5 w ^{3} - w}{w}

\frac{d}{d x} ((x^{2} - 3)^{4})

What is (\partial)/(\partial x)(5ln(1+x^2-y)) ?
The answer to (\partial)/(\partial x)(5ln(1+x^2-y)) is (10x)/(1+x^2-y)