Question

For the given​ functions, ​(a) express StartFraction dw Over dt EndFraction as a function of​ t, both by using the chain rule and by expressing w in terms of t and differentiating directly with respect to t. Then ​(b) evaluate StartFraction dw Over dt EndFraction at the given value of t. wequals6 y e Superscript x Baseline minus ln z​, xequalsln (t squared plus 1 )​, yequalstangent Superscript negative 1 Baseline t​, zequalse Superscript t​; tequals1

Answer 1

se explanation, and the final value is 16.995574

Step-by-step explanation:

a) As
`W` is a function of
`x(t), y(t), z(t)`. In order to use chain rule first we compute the partial derivatives of
`W` with respect to
`x(t), y(t), z(t)`.

`(\partial W )/(\partial x) = 7ye^(x)\\(\partial W )/(\partial y) = 7e^(x)\\(\partial W )/(\partial z) = -(1)/(z)`

then the total derivatives of each term with respect to t:

`(dx)/(dt) =(1)/(t^+1)*2t=(2t)/(t^2 + 1)\\(dy)/(dt) = (1)/(t^2+1) \\(dz)/(dt) = e^t\\\\`

we get:

`(dW)/(dt)= (\partial W )/(\partial x)(dx)/(dt) +(\partial W )/(\partial y)(dy)/(dt) +(\partial W )/(\partial z)(dz)/(dt)\\(dW)/(dt) = 7ye^(x)(2t)/(t^2 + 1) + 7e^(x)(1)/(t^2+1)-(1)/(z)e^t`

next we replace x,y and z with the respective functions:

`(dW)/(dt) = 7(tan^(-1)(t))e^(ln(t^2+1))(2t)/(t^2 + 1) + 7e^(ln(t^2+1))(1)/(t^2+1)-(1)/(e^t)e^t`

we cancel out the terms at the end, e^ln(expression) always will be equal to expression

after some algebra we get:

`(dW)/(dt) = 7(tan^(-1)(t))e^(ln(t^2+1))(2t)/(t^2 + 1) + 7e^(ln(t^2+1))(1)/(t^2+1)-(1)/(e^t)e^t\\(dW)/(dt) = 14t(tan^(-1)(t))+6`

b) as a function of t we directly replace
`x(t), y(t), z(t)`. and derivate with respect to t, before derivating, we can cancel out some terms:

`(dw)/(dt)= (d)/(dt) (7(tan^(-1)(t))*e^(ln(t^2+1))-ln(e^(t)))\\(dw)/(dt)= (d)/(dt) (7(tan^(-1)(t))*(t^2+1)-t)= (d)/(dt)( 7t^2*tan^(-1)(t)+7tan^(-1)(t)-t)\\ (dw)/(dt)=14t*tan^(-1)(t) + 7t^2*(1)/(t^2+1)+ 7(1)/(t^2+1) -1\\(dw)/(dt)=14t*tan^(-1)(t) + 7(t^2+1)*(1)/(t^2+1) -1= 14t*tan^(-1)(t)+6`

now we replace t = 1

`(dw)/(dt)(t=1)= 14*tan^(-1)(1)+6 = 6 +(7\pi )/(2) = 16.995574`