Differentiating related function

Calculus Tutorial

Differentiating related function

Intro

d/dt(sinx(t))=d/dx(sinx).dx/dt
d/dt(cosy(t))=d/dy(cosy).dy/dt

Sample Problem

The differentiable function x and y are related by the following equation
sin(x)+cos(y)=1 also dx/dt=7
find dy/dt when y=π/3

Solution

Sinx+cosy=1
d/dt(sinx(t)+d/dt(cos(y)=d/dt(1) — differentiate
d/dx(sinx).dx/dt+d/dy(cosy).dy/dt=0 — chain rule
cosx.dx/dt-siny.dy/dt=0
dx/dt=7 and y=π/3 — given
sinx+cosy=1
sinx+cosπ/3=1
sinx+1/2=1 — cosπ/3=1/2
sinx=1/2 — sinπ/6=1/2
x=π/6
cosx.dx/dt-siny.dy/dt=0
cosπ/6.7-sinπ/3.dy/dt=o — substitute
7. √3/2- √3/2.dy/dt=0
Dy/dt=7



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I was a teacher in Srilanka from 1992 to 2006. I took tuition from 1990 to 1991 in India. I do general work in Canada but I help to do homework to my daughters. I finished my teacher\'s training college diploma in Srilanka. I completed my B.Sc degree course at Bharayiar university in India.
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