3. Let: u=es+t and {x3t+x2t−4es−x2s3+1==00(A)(B)}Find: dxdu
soroban said:Hello, chengeto!
Your formula is correct . . . your partial differentiation isn't.
3. Let: u=es+t and {x3t+x2t−4es−x2s3+1==00(A)(B)}Find: dxdu
Formula: dxdu=∂s∂u⋅dxds+∂t∂u⋅dxdt .[1]
\(\displaystyle \text{We have: }\:\boxed{\begin{array}{ccc}\dfrac{\partial u}{\partial s} &=& e^{s+t} \\ \\[-3mm] \dfrac{\partial u}{\partial t} &=& e^{s+t} \end{array}}\) .[2]
Differentiate (A) w.r.t. x:x3dxdt+3x2+x2dxdt+2xt=0⇒x3dxdt+x2dxdt=−(3x2+2xt)
. . . . . . . . . . . . . . . . . . . x2(x+1)dxdt=−x(3x+2t)⇒dxdt=−x(x+1)3x+2t .[3]
Differentiate (B) w.r.t. x:esdxds−3x2s2dxds−2xs3=0⇒esdxds−3x2s2dxds=2xs3
. . . . . . . . . . . . . . . . . . . . . . . . (ex−3x2s2)dxds=2xs3⇒dxds=es−3x2s22xs2 .[4]
Substitute [2], [3], [4] into [1]:
. . dxdu=(es+t)[es−3x2s22xs2]+(es+t)[−x(x+1)3x+2t]=es+t[es−3x2s22xs2−x(x+1)3x+2t]