Question

Differentiate in two ways, using product rule and otherwise, the function (1+2 tanx) (5+4 cosx). Verify taht the answers are the same.

Answer 1

Let u=1+2 tanx , v= 5+4 cosx
Then $u^{\prime }=2se{c}^{2}x,v^{\prime }=-4sinx$
Using product rule:
$\frac{d}{dx}\left(uv\right)=u{v}^{\prime }+v{u}^{\prime }$
$\frac{d}{dx}(1+2tanx)(5+4cosx)$
$=(1+2tanx)(-4sinx)+(5+4cos‘x)\left(2se{c}^{2}x\right)$
$=-4sinx-8tanxsinx+10se{c}^{2}x+8secx$
$=-4sinx+10se{c}^{2}x(\frac{8}{cosx}-\frac{8si{n}^{2}x}{cosx})$
$=-4sinx+10se{c}^{2}x+8\left(\frac{co{s}^{2}x}{cosx}\right)$
$=-4sinx+10se{c}^{2}x+8cosx$
2nd method
$(1+2tanx)(5+4cosx)=5+4cosx+10tanx+8sinx$
Now, we have,
$\frac{d}{dx}\left[\right(1+2tanx\left)\right(5+4cosx\left)\right]$
$=\frac{d}{dx}[5+4cosx+10tanx+8sinx]=-4sinx+10se{c}^{2}x+8cosx$
Using both the methods, we get the same answer.