What is (\partial)/(\partial x)(ln(t^2+x^2)) ?

The answer to (\partial)/(\partial x)(ln(t^2+x^2)) is (2x)/(t^2+x^2)

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(\partial)/(\partial x)(ln(t^2+x^2))

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

\frac{2 x}{t ^{2} + x ^{2}}

Solution steps

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

Treat

t

as a constant

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{t ^{2} + x ^{2}} \cdot 2 x : \frac{2 x}{t ^{2} + x ^{2}}

= \frac{2 x}{t ^{2} + x ^{2}}

f (x) = 18 x

\int \frac{11}{x ^{3} - 125} d x

y = ln (6 - 7 x)

\frac{d}{d x} (cos^{2} (7 x))

(y + 4) y^{^{'}} = y (2 x + 4)

What is (\partial)/(\partial x)(ln(t^2+x^2)) ?
The answer to (\partial)/(\partial x)(ln(t^2+x^2)) is (2x)/(t^2+x^2)