Implicit Differentiation Practice

[ericndegwa]

Homework Statement

find dy/dx if x=1/1 + t and y = t2/1 + t

Homework Equations

dy/dx = (dy/dt) / (dx/dt)

The Attempt at a Solution

dx/dt = 1 * (1 + t )^-1

dx/dt = -1/ (1 +t)^2

dy/dt = t^2 * (1 +t)^-1

dy/dt = -2t/(1+ t)^2

therfore dy/dx = -2t/(1 + t)^2 / -1/(1 + t)^2

2t

 

[DryRun]

Your equations aren't clear. Edit your post to fix them or just attach a picture/screenshot.

x=1/1 + t meaning x= 1 +t ? or x= 1/ (1+t) ? (hence the importance of using parentheses!)

y = t2/1 + t

dy/dt = t^2 * (1 +t)^-1 ? So, y = t^2/1 + t ?

 

[jedishrfu]

look at dy/dt again its not just 2 t/ (1+t) there's a term missing assuming that y=t^2 /(1+t)

 

[Mark44]

Homework Statement

find dy/dx if x=1/1 + t and y = t2/1 + t

Homework Equations

dy/dx = (dy/dt) / (dx/dt)

The Attempt at a Solution

dx/dt = 1 * (1 + t )^-1

dx/dt = -1/ (1 +t)^2

dy/dt = t^2 * (1 +t)^-1

dy/dt = -2t/(1+ t)^2

Aside from the missing parentheses that have already been mentioned, you are showing equations that are incorrect. The first and third equations above are x and y, respectively, not dx/dt and dy/dt.

In calculating dy/dt, you need to use either the product rule (on t²(1 + t)^-1) or the quotient rule (on t²/(1 + t) ).

therfore dy/dx = -2t/(1 + t)^2 / -1/(1 + t)^2

2t