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 Question from Erson, a student: Find y' of the given function: y = cos^3x.

Erson,

This is an exercise with using the chain rule. The key is to realize that cos3x is another way to write (cosx)3. Hence you differentiate cos3x much the same way you would differentiate (x2 - 7x + 9)3. So let's do it.

If y = (x2 - 7x + 9)3 then to find y' you first notice that y = (something)3 and you know how to differentiate something raised to the third power because if

y = x3 then y' = 3x2.

But my function, y = (x2 - 7x + 9)3 isn't quite that simple, it's y = (something)3 and the chain rule tells me that

y' = 3(x2 - 7x + 9)2 × (the derivative of x2 - 7x + 9).

But I know the derivative of x2 - 7x + 9, its 2x - 7. Thus what I have is that if

y = (x2 - 7x + 9)3

then

y' = 3(x2 - 7x + 9)2 × (the derivative of x2 - 7x + 9) = 3(x2 - 7x + 9)2 × (2x - 7).

Now you try this with y = cos3x = (cosx)3.

Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.