AuthorWrite something about yourself. No need to be fancy, just an overview. Archives
April 2018
Categories
All

Back to Blog
The following problem may require the use of the following basic trigonometry derivatives. Also, the derivative, d/dx, may be represented by D.
\[\frac{d}{dx}sinx = cosx\]
\[\frac{d}{dx}cosx = sinx\]
\[sin^{2}x + cos^{2}x = 1\]
Solution: Apply the Quotient rule.
\[\frac{df(x)}{dx} = \frac{(1 + sinx) D(cosx)  D(1 + sinx) cosx}{(1 + sinx)^2}\]
\[= \frac{(1 + sinx)(sinx)  (cosx)cosx)}{(1 + sinx)^2}\]
\[= \frac{sinx  sin^2x  cos^2x}{(1 + sinx)^2}\]
\[= \frac{sinx  (sin^2x + cos^2x)}{(1 + sinx)^2}\]
From sin^2x + cos^2x = 1, we get
\[= \frac{sinx  1}{(1 + sinx)^2}\]
\[= \frac{(1 + sinx)}{(1 + sinx)^2}\]
\[= \frac{1}{(1 + sinx)}\]
0 Comments
read more
Leave a Reply. 