Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 957: 11

$dw/dt|_{t=1}= 3264$

Given that, $w=x^{3}y^{2}z^{4} ; x=t^{2},y=t+2,z=2t^{4}$ Calculate the partial derivatives $\frac{∂w}{∂x},\frac{∂w}{∂y},\frac{∂w}{∂z}$ $\frac{∂w}{∂x}=\frac{∂}{∂x}[x^{3}y^{2}z^{4}]$[Considering y and z as constants] $\frac{∂w}{∂x}=3x^{2}y^{2}z^{4}$ $\frac{∂w}{∂y}=\frac{∂}{∂x}[x^{3}y^{2}z^{4}]$[Considering x and z as constants] $\frac{∂w}{∂y}=2x^{3}yz^{4}$ $\frac{∂w}{∂z}=\frac{∂}{∂x}[x^{3}y^{2}z^{4}]$[Considering x and y as constants] $\frac{∂w}{∂z}=4x^{3}y^{2}z^{3}$ Calculate the ordinary derivatives $\frac{dw}{dx},\frac{dw}{dy},\frac{dw}{dz}$ $\frac{dx}{dt}=\frac{d}{dt}[t^{2}]$ $\frac{dx}{dt}=2t$ $\frac{dy}{dt}=\frac{d}{dt}[t+2]$ $\frac{dy}{dt}=1$ $\frac{dz}{dt}=\frac{d}{dz}[2t^{4}]$ $\frac{dz}{dt}=8t^{3}$ According to the chain rule for derivatives $\frac{dw}{dt} = \frac{∂w}{∂x}\frac{dx}{dt}+\frac{∂w}{∂y}\frac{dy}{dt}+\frac{∂w}{∂z}\frac{dz}{dt}$ $\frac{dw}{dt} =(3x^{2}y^{2}z^{4})(2t)+(2x^{3}yz^{4})(1)+(4x^{3}y^{2}z^{3})(8t^{3})$ $\frac{dw}{dt} =6x^{2}y^{2}z^{4}t+2x^{3}yz^{4}+32x^{3}y^{2}z^{3}t^{3}$ Here $x=t^{2},y=t+2,z=2t^{4}$. So, $\frac{dw}{dt} =6(t^{2})^{2}(t+2)^{2}(2t^{4})^{4}t+2(t^{2})^{3}(t+2)(2t^{4})^{4}+32(t^{2})^{3}(t+2)^{2}(2t^{4})^{3}t^{3}$ $\frac{dw}{dt} =96t^{21}(t^{2}+4t+4)+32t^{22}(t+2)+256t^{21}(t^{2}+4t+4)$ $\frac{dw}{dt} =352t^{21}(t^{2}+4t+4)+32t^{22}(t+2)$ $\frac{dw}{dt} =352t^{23}+1408t^{22}+1408t^{21}+32t^{23}+64t^{22}$ $\frac{dw}{dt} =384t^{23}+1472t^{22}+1408t^{21}$ Calculate the rate of change of w with respect to t at t = 1 $dw/dt|_{t=1}=384(1)^{23}+1472(1)^{22}+1408(1)^{21}$ $dw/dt|_{t=1}=384+1472+1408$ $dw/dt|_{t=1}=3264$ Checking the work by expressing w as a function of t and differentiating. $w = f(x,y,z) =x^{3}y^{2}z^{4}$ $w = f(t) = (t^{2})^{3}(t+2)^{2}(2t^{4})^{4}$ $w = f(t) = 16t^{22}(t^{2}+4t+4)$ $w = f(t) = 16t^{24}+64t^{23}+64t^{22}$ Calculate $\frac{dw}{dt}$ $\frac{dw}{dt}=\frac{d}{dt}[16t^{24}+64t^{23}+64t^{22}]$ $\frac{dw}{dt}= (16)(24)t^{23}+(64)(23)t^{22}+(64)(22)t^{21}$ $\frac{dw}{dt}= 384t^{23}+1472t^{22}+1408t^{21}$ Calculate $\frac{dw}{dt}$ at t = 1 $dw/dt|_{t=1}=384(1)^{23}+1472(1)^{22}+1408(1)^{21}$ $dw/dt|_{t=1}=384+1472+1408$ $dw/dt|_{t=1}=3264$