What is the derivative of sqrt(s^2+t^2) ?

The derivative of sqrt(s^2+t^2) is s/(sqrt(s^2+t^2))

What is the first derivative of sqrt(s^2+t^2) ?

The first derivative of sqrt(s^2+t^2) is s/(sqrt(s^2+t^2))

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derivative of

s^{2} + t^{2}

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

Solution steps

\frac{d}{d s} (s^{2} + t^{2})

Treat

t

as a constant

Apply the chain rule:

\frac{1}{2 s ^{2} + t ^{2}} \frac{d}{d s} (s^{2} + t^{2})

= \frac{1}{2 s ^{2} + t ^{2}} \frac{d}{d s} (s^{2} + t^{2})

\frac{d}{d s} (s^{2} + t^{2}) = 2 s

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

Simplify

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

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

\int e^{c o s (π x)} sin (π x) d x

n = 0 \sum \infty (n) sin (\frac{1}{n})

y^{^{'}} + a y - b = 0

\int \frac{5}{4 + 9 r ^{2}} d r

x \to \infty lim (4 + x)

What is the derivative of sqrt(s^2+t^2) ?
The derivative of sqrt(s^2+t^2) is s/(sqrt(s^2+t^2))
What is the first derivative of sqrt(s^2+t^2) ?
The first derivative of sqrt(s^2+t^2) is s/(sqrt(s^2+t^2))