Exercises — Chain rule for multivariable functions — all cases

$t$ tak do paths hain: ek $x$ ke through, ek $y$ ke through. Har ek ke saath multiply karo, add karo: $\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}.$ Left side par $d$ likhte hain kyunki $x(t),y(t)$ substitute karne ke baad output $z$ sirf ek ultimate variable $t$ par depend karta hai. Andar ke do pieces $\partial$ rehte hain kyunki $z$ ke genuinely do inputs $x,y$ hain.

Teen inner functions $\Rightarrow$ $w$ se $t$ tak teen paths: $\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}.$ Saare symbols $\partial$ hain: output do ultimate variables $s$ aur $t$ par depend karta rahe‍ga.

Ab $t$ ke teen raaste hain: $x$ ke through, $y$ ke through, aur $f$ mein seedha explicit slot ke through: $\frac{dw}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial t}.$ Aakhri term $\partial f/\partial t$ ka matlab hai "$x,y$ ko fixed rakho, sirf explicit $t$ ko wiggle karo." Ise miss karna classic galti hai.

Link derivatives. $\frac{\partial z}{\partial x}=2xy$, $\frac{\partial z}{\partial y}=x^2+3$. Phir $\frac{dx}{dt}=2t$, $\frac{dy}{dt}=e^t$. Assemble (do paths, add): $\frac{dz}{dt}=2xy\cdot 2t+(x^2+3)e^t.$ Substitute $x=t^2,y=e^t$: $\frac{dz}{dt}=2t^2{e}^t\cdot 2t+(t^4+3)e^t=4t^3{e}^t+(t^4+3)e^t.$ $t=0$ par: $x=0,y=1$, isliye $\frac{dz}{dt}{|}_0=0+(0+3)\cdot 1=\boxed{3}.$

$\frac{\partial z}{\partial x}=\frac{2x}{x^2+y^2}$, $\frac{\partial z}{\partial y}=\frac{2y}{x^2+y^2}$. Aur $\frac{\partial x}{\partial \theta }=-r\sin \theta$, $\frac{\partial y}{\partial \theta }=r\cos \theta$. $\frac{\partial z}{\partial \theta }=\frac{2x}{x^2+y^2}(-r\sin \theta )+\frac{2y}{x^2+y^2}(r\cos \theta ).$ Kyunki $x^2+y^2=r^2$ hai, $x=r\cos \theta ,y=r\sin \theta$ substitute karo: $=\frac{2r\cos \theta (-r\sin \theta )+2r\sin \theta (r\cos \theta )}{r^2}=\frac{-2r^2\cos \theta \sin \theta +2r^2\sin \theta \cos \theta }{r^2}=\boxed{0}.$ Geometry: $z=\ln (r^2)$ sirf distance $r$ par depend karta hai, angle par nahi — origin ke around rotate karne se $z$ nahi badalti, isliye uski $\theta$-rate zero hai. Rule yeh fact "discover" karta hai.

$\frac{\partial w}{\partial x}=y,\frac{\partial w}{\partial y}=x,\frac{\partial w}{\partial z}=1$. Inner rates: $1,2t,3t^2$. $\frac{dw}{dt}=y\cdot 1+x\cdot 2t+1\cdot 3t^2=t^2+t\cdot 2t+3t^2=t^2+2t^2+3t^2=6t^2.$ $t=2$ par: $6(4)=\boxed{24}$. (Check: $w=t\cdot t^2+t^3=2t^3$, isliye $dw/dt=6t^2$. ✓)

Yeh "implicit reappearance" case hai — $t$ ke teen paths hain. $\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial y}=2y,\frac{\partial f}{\partial t}=1$. Inner: $\frac{dx}{dt}=-\sin t,\frac{dy}{dt}=\cos t$. $\frac{dw}{dt}=2x(-\sin t)+2y(\cos t)+1.$ $x=\cos t,y=\sin t$ substitute karo: $=-2\cos t\sin t+2\sin t\cos t+1=0+1=\boxed{1}.$ Interpretation: point $(x,y)$ unit circle ke around ghoomta hai, isliye $x^2+y^2=1$ kabhi nahi badalta — woh do paths cancel ho jaate hain. Sirf explicit $+t$ term bachti hai, jo rate $1$ deti hai. $\partial f/\partial t$ chhod do aur galat answer $0$ milega.

$\frac{\partial u}{\partial x}=1,\frac{\partial v}{\partial x}=-1$, aur $\frac{\partial u}{\partial y}=-1,\frac{\partial v}{\partial y}=1$. $\frac{\partial z}{\partial x}=f_u(1)+f_v(-1)=f_u-f_v,$ $\frac{\partial z}{\partial y}=f_u(-1)+f_v(1)=-f_u+f_v.$ Add karo: $(f_u-f_v)+(-f_u+f_v)=\boxed{0}$. Kyun: $v=-u$ hai, isliye $z$ actually ek combination $x-y$ par depend karta hai; $x$ aur $y$ ko equal amounts se move karne par $x-y$ fixed rehta hai, isliye koi change nahi hota.

Inner rates: $x_s=e^s\cos t=x$, $y_s=e^s\sin t=y$, $x_t=-e^s\sin t=-y$, $y_t=e^s\cos t=x$. $z_s=z_{x}x+z_{y}y,z_t=z_x(-y)+z_{y}x=-z_{x}y+z_{y}x.$ Square karke add karo: $z_s^2+z_t^2=(z_{x}x+z_{y}y{)}^2+(-z_{x}y+z_{y}x{)}^2.$ Expand karo: $z_x^2{x}^2+2z_x{z}_{y}xy+z_y^2{y}^2+z_x^2{y}^2-2z_x{z}_{y}xy+z_y^2{x}^2=z_x^2(x^2+y^2)+z_y^2(x^2+y^2)$. Kyunki $x^2+y^2=e^{2s}$ hai: $\boxed{z_s^2+z_t^2=e^{2s}\left(z_x^2+z_y^2\right)}.$

Pure bookkeeping — $f$ ka koi formula nahi chahiye: $\frac{dz}{dt}=f_x{x}_t+f_y{y}_t=(3)(2)+(-1)(5)=6-5=\boxed{1}.$

Inner rates: $x_s=1,y_s=1,x_t=1,y_t=-1$. $z_s=z_x(1)+z_y(1)=z_x+z_y,z_t=z_x(1)+z_y(-1)=z_x-z_y.$ Multiply karo — difference of squares: $z_s{z}_t=(z_x+z_y)(z_x-z_y)=z_x^2-z_y^2.\boxed{✓}$

First derivative (parent note se): $z_r=f_x\cos \theta +f_y\sin \theta$. Phir se $r$ ke w.r.t. differentiate karo, $\theta$ ko fixed rakho. Trig factors $\cos \theta ,\sin \theta$ $r$ mein constants hain; lekin $f_x$ aur $f_y$ ab bhi $x,y$ ke functions hain, isliye $r$ ke bhi. Har ek par chain rule lagao: $\frac{\partial f_x}{\partial r}=f_{xx}\cos \theta +f_{xy}\sin \theta ,\frac{\partial f_y}{\partial r}=f_{yx}\cos \theta +f_{yy}\sin \theta .$ Toh $z_{rr}=\cos \theta (f_{xx}\cos \theta +f_{xy}\sin \theta )+\sin \theta (f_{xy}\cos \theta +f_{yy}\sin \theta ),$ $f_{yx}=f_{xy}$ use karte hue (equality of mixed partials). Collect karo: $\boxed{z_{rr}=f_{xx}{\cos }^2\theta +2f_{xy}\cos \theta \sin \theta +f_{yy}{\sin }^2\theta }.$

$u=\lambda x$, $v=\lambda y$ lo, aur $g(\lambda )=f(u,v)=f(\lambda x,\lambda y)$ define karo. Dono sides ko $\lambda$ mein differentiate karo. Left side mein chain rule se (do paths, $u$ aur $v$ ke through): $g^{\prime }(\lambda )=f_u\frac{du}{d\lambda }+f_v\frac{dv}{d\lambda }=f_{u}x+f_{v}y.$ Right side: $\frac{d}{d\lambda }({\lambda }^{k}f(x,y))=k{\lambda }^{k-1}f(x,y).$ Yeh har $\lambda$ ke liye equal hain. $\lambda =1$ set karo, toh $u=x,v=y$: $x{f}_x+y{f}_y=kf.\boxed{✓}$ Trick yeh hai: scalar knob $\lambda$ introduce karo, uske through differentiate karo, phir $1$ par freeze karo.

Chaar outputs mein se har ek par scalar chain rule apply karo: $u_s=f_x{x}_s+f_y{y}_s,u_t=f_x{x}_t+f_y{y}_t,$ $w_s=g_x{x}_s+g_y{y}_s,w_t=g_x{x}_t+g_y{y}_t.$ Ab right-hand product ko entry-by-entry padho. $J_f$ ki row 1 aur $J_x$ ki column 1 ka dot product $f_x{x}_s+f_y{y}_s=u_s$ deta hai; column 2 $u_t$ deta hai; row 2 $w_s,w_t$ deti hai. Har entry match karti hai, isliye $J_{(u,w)}=J_f{J}_x.\boxed{✓}$ Isliye parent note general chain rule ko just matrix multiplication of Jacobians kehta hai — Gradient and Jacobian dekho.

$(s,t)=(1,2)$ par: $x=3,y=2$. Jacobians. $J_f=\left(\begin{array}{cc}2x& 1\\ y& x\end{array}\right)=\left(\begin{array}{cc}6& 1\\ 2& 3\end{array}\right)$. $J_x=\left(\begin{array}{cc}{x}_s& x_t\\ y_s& y_t\end{array}\right)=\left(\begin{array}{cc}1& 1\\ t& s\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right).$ Product top-left entry ($u_s$): $6\cdot 1+1\cdot 2=8$. Direct: $u=x^2+y=(s+t{)}^2+st$, isliye $u_s=2(s+t)+t=2(3)+2=\boxed{8}$. ✓