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Notice that (sin(x))'= cos(x) and (x^2)' = 2x

We use the product rule to differentiate, by noticing the expression is a product. 

so (fxgx)' = f'xgx + fx*g'x

substituting in ...

Substitute y into 4y-2=3x

4(2x-7)-2=3x   ->   8x-28-2=3x     ->      8x-3x=28+2    ->   5x=30    ->    x=30/5    ->    x=6

subsitute x into y=2x-7

y=2(6)-7    ->    ...

dy/dx = (4cos(2x) + 6sin(2x) + 2)/(2+cos(2x))²

First, write y in terms of cos(x). We are familiar with cos(x) and know how to differentiate it. We know that sec(x) = 1/cos(x) = (cos(x))^-1.  Next, find dy/dx in terms of cos(x) and sin(x). Ag...

In order to solve this inequality, it is helpful to have all the terms on the left-hand side of the equation. To do this, we can subtract 3(x + 6) from both sides of the equation. This gives us:

x^...