View Full Version : Derivatives involving a trig function
ankara55t
Sep 17, 2009, 04:28 PM
What are the steps behind differentiating and simplifying the following:
y = sin^3x
In getting y' = 3sin^2xcosx why do we not get cos^2x?
galactus
Sep 18, 2009, 03:50 AM
By using the chain rule.
\frac{d}{dx}[sin^{3}(x)]=3sin^{2}(x)cos(x)
The derivative of the inside times the derivative of the outside.
The derivative of the outside is 3sin^{2}(x)
The derivative of the inside is cos(x)
We get the cos(x) by just differentiating sin(x).
Unknown008
Sep 18, 2009, 10:33 AM
You can also have it with the product rule.
You know that \frac{d(sin^2(x))}{dx} = 2sin(x)cos(x) ?
Then put it like that: sin^3(x) = sin^2(x)sin(x)
Use the product rule here;
\frac{d(sin^2(x)sin(x))}{dx} = sin^2(x)cos(x) + sin(x).2sin(x)cos(x) = sin^2(x)cos(x) + 2sin^2(x)cos(x) = 3sin^2(x)cos(x)