Worked examples: Example 2
Cont’d from “Definition”
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Let’s use an outcome we already know. If our original
function of y
has a derivative with respect to y
= cos y
Now let’s calculate the derivative of f(y)
with respect to x.
Here we have to again apply implicit differentiation since y is a function of x
d/dxsin(y(x)) = d/dysin(y(x)) d/dxy(x) = cos(y(x)) dy/dx
Now we can simply substitute these into the equation determined previously.
dy/dx= dy/dxcos y [cos y]−1
= dy/dxcos y
since cosy/cos y= 1 the equation becomes
confirming the equation from “Implicit differentiation of equations” and, in turn, confirming the differentiation process used.