Use the chain rule to find ∂z ∂s and ∂z ∂t . z = tan−1(x6 + y6), x = s ln(t), y = tes ∂z ∂s = ∂z ∂t =

Answered question

2021-11-03

Use the chain rule to find 

z
s

 and 

z
t

.

z = tan−1(x6 + y6),   x = s ln(t),   y = tes

z
s

=

 

z
t

=


Answer & Explanation

user_27qwe

user_27qwe

Skilled2023-04-21Added 375 answers

To use the chain rule to find dzds and dzdt, we need to first express z as a function of s and t using the given equations for x and y.

Starting with x=sln(t), we can solve for s in terms of x and t:

x=sln(t)
s=xln(t)
 

Next, using y=tes, we can solve for t in terms of y and s:

y=tes
t=yes
 

Substituting these expressions into the equation for z, we get:

z=tan-1(x6+y6)
z=tan-1((xln(t))6+(yes)6)
 

Now we can differentiate z with respect to s and t using the chain rule.

To find dzds, we treat t as a constant and differentiate z with respect to s:

dzds=dds[tan-1((xln(t))6+(yes)6)]
     =(1(xln(t))6+(yes)6+1)(6(xln(t))5(1ln(t))(1t)es+6(yes)5(tes))
     =(6x5tln(t)(x2+y2))es+(6ty5e7(x2+y2))
 

Similarly, to find dzdt, we treat s as a constant and differentiate z with respect to t:


dzdt=ddt[tan-1((xln(t))6+(yes)6)]
     =(1(xln(t))6+(yes)6+1)(6(xln(t))5(1x)(-tln2(t))es+6(yes)5(-ses))
     =-(6tx5ln2(t)(x2+y2))es-(6sy5e8(x2+y2))  
 

Therefore, dzds=(6x5tln(t)(x2+y2))es+(6ty5e7(x2+y2)) and dzdt=-(6tx5ln2(t)(x2+y2))es-(6sy5e8(x2+y2)).

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