Find the first partial derivatives of the function. w=evu+v2

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komunidadO · Accepted Answer

We havew=evu+v2&amp;nbsp;Then,wu=(0)(u+v2)−ev(1)(u+v2)2&amp;nbsp;wu=−ev(u+v2)2&amp;nbsp;wv=(ev)(u+v2)−(ev)(2v)(u+v2)2&amp;nbsp;wv=(u+v2−2v)ev(u+v2)2&amp;nbsp;wu=−ev(u+v2)2&amp;nbsp;wv=(u+v2−2v)ev(u+v2)2

madeleinejames20 · Accepted Answer

First, let&#039;s find ∂w∂u:Using the quotient rule, we have:∂w∂u=∂∂u(ev)·(u+v2)−(ev)·∂∂u(u+v2)(u+v2)2Now, let&#039;s calculate each partial derivative:∂∂u(ev)=0since v does not contain u.Next,∂∂u(u+v2)=1as u is being differentiated with respect to u, and v2 does not contain u.Substituting these results back into the expression for ∂w∂u, we get:∂w∂u=−ev(u+v2)2Now, let&#039;s find ∂w∂v:Again using the quotient rule, we have:∂w∂v=∂∂v(ev)·(u+v2)−(ev)·∂∂v(u+v2)(u+v2)2Now, let&#039;s calculate each partial derivative:∂∂v(ev)=evsince v is being differentiated with respect to v, and u does not contain v.Next,∂∂v(u+v2)=2vas v2 is being differentiated with respect to v, and u does not contain v.Substituting these results back into the expression for ∂w∂v, we get:∂w∂v=ev·(u+v2)−(ev)·2v(u+v2)2Therefore, the first partial derivatives of the function w are:∂w∂u=−ev(u+v2)2and∂w∂v=ev·(u+v2)−(ev)·2v(u+v2)2

Mr Solver · Accepted Answer

Step 1:To find the first partial derivatives of the function, w=evu+v2, with respect to u and v, we can proceed as follows:First, let&#039;s find the partial derivative of w with respect to u, denoted as ∂w∂u. To do this, we treat v as a constant and differentiate w with respect to u. Using the quotient rule, we have:∂w∂u=(∂∂u(ev))(u+v2)−(ev)(∂∂u(u+v2))(u+v2)2The derivative of ev with respect to u is 0 since v is not a function of u. The derivative of u+v2 with respect to u is simply 1. Simplifying the expression, we get:∂w∂u=−ev(u+v2)2Step 2:Now, let&#039;s find the partial derivative of w with respect to v, denoted as ∂w∂v. This time, we treat u as a constant and differentiate w with respect to v. Again, using the quotient rule, we have:∂w∂v=(ev)(∂∂v(u+v2))−(∂∂v(ev))(u+v2)(u+v2)2The derivative of u+v2 with respect to v is simply 2v. The derivative of ev with respect to v is ev. Simplifying the expression further, we get:∂w∂v=ev(2v)−(ev)(u+v2)(u+v2)2=2vev−(u+v2)ev(u+v2)2Hence, the first partial derivatives of the function are:∂w∂u=−ev(u+v2)2∂w∂v=2vev−(u+v2)ev(u+v2)2

Eliza Beth13 · Accepted Answer

Result:∂w∂u=−ev(u+v2)2∂w∂v=ev(u+v2−2v)(u+v2)2Solution:∂w∂u=∂∂u(evu+v2)To differentiate w with respect to u, we can treat v as a constant. Applying the quotient rule, we have:∂w∂u=(v2)·0−ev·1(u+v2)2=−ev(u+v2)2Now, let&#039;s find the partial derivative with respect to v:∂w∂v=∂∂v(evu+v2)To differentiate w with respect to v, we can treat u as a constant. Applying the quotient rule again, we get:∂w∂v=(u+v2)·ev−ev·2v(u+v2)2=ev(u+v2−2v)(u+v2)2Therefore, the first partial derivatives are:∂w∂u=−ev(u+v2)2∂w∂v=ev(u+v2−2v)(u+v2)2

Find the first partial derivatives of the function. w=e^{v}/u+v^{2}

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