Problem 21 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$G(s, t)=\frac{\sqrt{s t}}{s+t}$$

Step-by-Step Solution

Answer

Question: Find the first partial derivatives G_s(s, t) and G_t(s, t) of the function G(s, t) = (sqrt(st)) / (s + t). Answer: The first partial derivatives of the function G(s, t) are: G_s(s, t) = ( (sqrt(t) / (2*sqrt(s))) * (s + t) - (sqrt(st)) ) / (s + t)^2 G_t(s, t) = ( (sqrt(s) / (2*sqrt(t))) * (s + t) - (sqrt(st)) ) / (s + t)^2

Step 1: Find the partial derivative with respect to s

To find the partial derivative with respect to s, we use the quotient rule: $$\frac{d}{ds}\left(\frac{u(s, t)}{v(s, t)}\right) = \frac{u'_s v - u v'_s}{v^2}$$ Here, $$u(s, t) = \sqrt{s t} = (st)^{1/2}$$ and $$v(s, t) = s + t$$. Now, we differentiate u and v with respect to s: $$u'_s = \frac{1}{2}(st)^{-1/2} (t) = \frac{\sqrt{t}}{2\sqrt{s}}$$ $$v'_s = 1$$ Now, we apply the quotient rule: $$G_s(s, t) = \frac{u'_s v - u v'_s}{v^2} = \frac{\frac{\sqrt{t}}{2\sqrt{s}}(s+t) - (st)^{1/2}(1)}{(s+t)^2}$$

Step 2: Find the partial derivative with respect to t

To find the partial derivative with respect to t, we use the quotient rule: $$\frac{d}{dt}\left(\frac{u(s, t)}{v(s, t)}\right) = \frac{u'_t v - u v'_t}{v^2}$$ We have already differentiated u and v with respect to s. Now we need to differentiate u and v with respect to t: $$u'_t = \frac{1}{2}(st)^{-1/2} (s) = \frac{\sqrt{s}}{2\sqrt{t}}$$ $$v'_t = 1$$ Now, we apply the quotient rule: $$G_t(s, t) = \frac{u'_t v - u v'_t}{v^2} = \frac{\frac{\sqrt{s}}{2\sqrt{t}}(s+t) - (st)^{1/2}(1)}{(s+t)^2}$$ So the first partial derivatives of the function G(s, t) are: $$G_s(s, t) = \frac{\frac{\sqrt{t}}{2\sqrt{s}}(s+t) - (st)^{1/2}(1)}{(s+t)^2}$$ $$G_t(s, t) = \frac{\frac{\sqrt{s}}{2\sqrt{t}}(s+t) - (st)^{1/2}(1)}{(s+t)^2}$$