Calculate the derivatives using implicit differentiation...

Calculate the derivatives using implicit differentiation, where c is a constant: $\frac{\partial U}{\partial T}$ and $\frac{\partial T}{\partial U}$ of $(TU-V)^2 \ln(W-UV) = c$ at $(T, U, V, W) = (7, 1, 9, 18)$ $\frac{\partial U}{\partial T}|_{(7,1,9,18)} = \boxed{}$ $\frac{\partial T}{\partial U}|_{(7,1,9,18)} = \boxed{}$

Transcript

00:01 Hi, here in the given question we are given that we need to calculate the value of del u upon del t and del t upon del u.

00:10 Now further we are given that differentiation of tu minus v whole square multiplied with ln of w minus uv is equal to ln of phi and here we need to calculate this value at t comma u comma v comma w is equal to 1 comma 4 comma 5 comma 25.

00:29 So here in our case now we know that if we start to differentiate the above function with respect to t so del u upon del t can be further written as here we have 2 times of tu minus v multiplied with t into del u upon del t plus u multiplied with ln of w minus uv plus here in our case further we have tu minus v whole square divided by w minus uv and here this value is multiplied with v and the whole value again multiplied with del u upon del t is equal to 0.

01:14 So here in our case now we know that if we solve the value further and simplify this equation then this equation can be written as del u upon del t is equal to 2t multiplied with tu minus v ln of w minus uv minus v into tu minus v whole square divided by w minus uv and this is equal to 2u multiplied with v minus tu ln of w minus uv.

01:45 So on solving further we have value of del u upon del t is equal to 2v multiplied with v minus tu multiplied with ln of w minus uv and the whole value is divided by 2t into tu minus v multiplied with ln of w minus uv minus v multiplied with tu minus v whole square divided by w minus uv.

02:12 So this is the value.

02:13 Now further we need to substitute this value at a given point.

02:17 So the point given to us is at 2, 1, 4, 5, 25.

02:26 So substituting this value further we have.

02:31 So now substituting the value here in our case we have del u upon del t is equal to 2 multiplied with 4 multiplied with 5 minus 4 multiplied with ln of 25 minus 20 the whole value divided by 2 multiplied with 1 multiplied with 4 minus 5 multiplied with ln of 25 minus 20 minus 5 multiplied with 4 into 1 minus 5 whole square divided by 25 minus 20...

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