Implicit differentiation - Multivariable calculus

chengeto

New member
Joined
Feb 28, 2009
Messages
49
Guys l just want to check if this is correct ?
 

Attachments

  • Capture.JPG
    Capture.JPG
    74.7 KB · Views: 236
Hello, chengeto!

Your formula is correct . . . your partial differentiation isn't.


3. Let: u  =  es+t and {x3t+x2t4=0(A)esx2s3+1=0(B)}Find: dudx\displaystyle \text{3. Let: }\:u \;=\;e^{s+t}\:\text{ and }\:\begin{Bmatrix}x^3t + x^2t -4 &=& 0 & (A) \\ e^s - x^2s^3 + 1 &=& 0 & (B) \end{Bmatrix} \qquad \text{Find: }\:\frac{du}{dx}

Formula:   dudx  =  usdsdx+utdtdx\displaystyle \text{Formula: }\;\frac{du}{dx} \;=\;\frac{\partial u}{\partial s}\cdot\frac{ds}{dx} + \frac{\partial u}{\partial t}\cdot\frac{dt}{dx} .[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 ⁣:x3dtdx  +  3x2  +  x2dtdx  +  2xt=0x3dtdx  +  x2dtdx  =  (3x2  +  2xt)\displaystyle \text{Differentiate (A) w.r.t. }x\!:\quad x^3\frac{dt}{dx} \;+\; 3x^2\;+\; x^2\frac{dt}{dx} \;+\; 2xt\:=\:0 \quad\Rightarrow\quad x^3\frac{dt}{dx} \;+\; x^2\frac{dt}{dx} \;=\;-(3x^2 \;+\; 2xt)

. . . . . . . . . . . . . . . . . . . x2(x+1)dtdx  =  x(3x+2t)dtdx  =  3x+2tx(x+1)\displaystyle x^2(x+1)\frac{dt}{dx} \;=\;-x(3x+2t) \quad\Rightarrow\quad \boxed{\frac{dt}{dx} \;=\;-\frac{3x+2t}{x(x+1)}} .[3]


Differentiate (B) w.r.t. x ⁣:esdsdx    3x2s2dsdx    2xs3    =    0esdsdx    3x2s2dsdx    =    2xs3\displaystyle \text{Differentiate (B) w.r.t. }x\!:\quad e^s\frac{ds}{dx}\;-\;3x^2s^2\frac{ds}{dx}\;-\;2xs^3 \;\;=\;\;0 \quad\Rightarrow\quad e^s\frac{ds}{dx}\;-\;3x^2s^2\frac{ds}{dx} \;\;=\;\;2xs^3

. . . . . . . . . . . . . . . . . . . . . . . . (ex3x2s2)dsdx  =  2xs3dsdx  =  2xs2es3x2s2\displaystyle \left(e^x-3x^2s^2\right)\frac{ds}{dx} \;=\;2xs^3 \quad\Rightarrow\quad \boxed{\frac{ds}{dx} \;=\;\frac{2xs^2}{e^s-3x^2s^2}} .[4]


Substitute [2], [3], [4] into [1]:

. . dudx    =    (es+t)[2xs2es3x2s2]+(es+t)[3x+2tx(x+1)]    =    es+t[2xs2es3x2s23x+2tx(x+1)]\displaystyle \frac{du}{dx}\;\;=\;\;\left(e^{s+t}\right)\left[\frac{2xs^2}{e^s-3x^2s^2}\right] + \left(e^{s+t}\right)\left[-\frac{3x+2t}{x(x+1)}\right] \;\;=\;\;e^{s+t}\left[\frac{2xs^2}{e^s-3x^2s^2} - \frac{3x+2t}{x(x+1)}\right]


 
soroban said:
Hello, chengeto!

Your formula is correct . . . your partial differentiation isn't.


3. Let: u  =  es+t and {x3t+x2t4=0(A)esx2s3+1=0(B)}Find: dudx\displaystyle \text{3. Let: }\:u \;=\;e^{s+t}\:\text{ and }\:\begin{Bmatrix}x^3t + x^2t -4 &=& 0 & (A) \\ e^s - x^2s^3 + 1 &=& 0 & (B) \end{Bmatrix} \qquad \text{Find: }\:\frac{du}{dx}

Formula:   dudx  =  usdsdx+utdtdx\displaystyle \text{Formula: }\;\frac{du}{dx} \;=\;\frac{\partial u}{\partial s}\cdot\frac{ds}{dx} + \frac{\partial u}{\partial t}\cdot\frac{dt}{dx} .[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 ⁣:x3dtdx  +  3x2  +  x2dtdx  +  2xt=0x3dtdx  +  x2dtdx  =  (3x2  +  2xt)\displaystyle \text{Differentiate (A) w.r.t. }x\!:\quad x^3\frac{dt}{dx} \;+\; 3x^2\;+\; x^2\frac{dt}{dx} \;+\; 2xt\:=\:0 \quad\Rightarrow\quad x^3\frac{dt}{dx} \;+\; x^2\frac{dt}{dx} \;=\;-(3x^2 \;+\; 2xt)

. . . . . . . . . . . . . . . . . . . x2(x+1)dtdx  =  x(3x+2t)dtdx  =  3x+2tx(x+1)\displaystyle x^2(x+1)\frac{dt}{dx} \;=\;-x(3x+2t) \quad\Rightarrow\quad \boxed{\frac{dt}{dx} \;=\;-\frac{3x+2t}{x(x+1)}} .[3]


Differentiate (B) w.r.t. x ⁣:esdsdx    3x2s2dsdx    2xs3    =    0esdsdx    3x2s2dsdx    =    2xs3\displaystyle \text{Differentiate (B) w.r.t. }x\!:\quad e^s\frac{ds}{dx}\;-\;3x^2s^2\frac{ds}{dx}\;-\;2xs^3 \;\;=\;\;0 \quad\Rightarrow\quad e^s\frac{ds}{dx}\;-\;3x^2s^2\frac{ds}{dx} \;\;=\;\;2xs^3

. . . . . . . . . . . . . . . . . . . . . . . . (ex3x2s2)dsdx  =  2xs3dsdx  =  2xs2es3x2s2\displaystyle \left(e^x-3x^2s^2\right)\frac{ds}{dx} \;=\;2xs^3 \quad\Rightarrow\quad \boxed{\frac{ds}{dx} \;=\;\frac{2xs^2}{e^s-3x^2s^2}} .[4]


Substitute [2], [3], [4] into [1]:

. . dudx    =    (es+t)[2xs2es3x2s2]+(es+t)[3x+2tx(x+1)]    =    es+t[2xs2es3x2s23x+2tx(x+1)]\displaystyle \frac{du}{dx}\;\;=\;\;\left(e^{s+t}\right)\left[\frac{2xs^2}{e^s-3x^2s^2}\right] + \left(e^{s+t}\right)\left[-\frac{3x+2t}{x(x+1)}\right] \;\;=\;\;e^{s+t}\left[\frac{2xs^2}{e^s-3x^2s^2} - \frac{3x+2t}{x(x+1)}\right]




Thanks
 
Top