Question

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.
$\left(i\right)(3x^2+2{)}^2$
$\left(ii\right)(x+2)(x+3)$
$\left(iii\right)(3secx-4cosecx)(-2sinx+5cosx)$

Answer 1

Let $u=3x^2+2,u=3x^2+2$
Then, $u^{\prime }=6x,v^{\prime }=6x$
Using the product rule
$\frac{d}{dx}\left(uv\right)=u{v}^{\prime }+v{u}^{\prime }$
$\frac{d}{dx}\left[\right(3x^2+2\left)\right(3x^2+2\left)\right]=\left(3x^2\right)\left(6x\right)+(3x^2+2)\left(6x\right)$
$\frac{d}{dx}\left[\right(3x^2\left)\right(3x^2+2\left)\right]=18x^3+12x+18x^3+12x$
$\frac{d}{dx}\left[\right(3x^2+2\left)\right(3x^2+2\left)\right]=36x^3+24x$
2nd method
$\frac{d}{dx}\left[\right(3x^2+2{)}^2]=\frac{d}{dx}(9x^4+12x^2+4)=36x^3+24x$
Using both methods, we get the same answer.
(ii) (x+2)(x+3)
Let u=x+2, v= x+3
Then, u'= 1, v'= 1
Using the product rule
$\frac{d}{dx}\left(uv\right)=u{v}^{\prime }+v{u}^{\prime }$
$\frac{d}{dx}\left[\right(x+2\left)\right(x+3\left)\right]$
$=(x+2)1+(x+3)1$
$=x+2+x+3$
$=2x+5$
2nd method
$\frac{d}{dx}\left[\right(x+2\left)\right(x+3\left)\right]=\frac{d}{dx}[x^2+5x+6]$
$=2x+5$
Using both the methods, we get the same answer.
(iii)$(3secx-4cosecx)(-2sinx+5cosx)$ Product rule (1st method)
Let $u=3secx-4cosecx,v=-2sinx+5cosx$
Then $u^{\prime }=3secxtanx+4cosecxcotx$
$v^{\prime }=-2cosx-5sinx$
Using the product rule
$\frac{d}{dx}\left(uv\right)=u{v}^{\prime }+v{u}^{\prime }$
$\frac{d}{dx}\left[\right(3secx-4cosecx\left)\right(-2sinx+5cosx)=(3secx-4cosec4\left)\right(-2cosx-5sinx)+(-2sinx+5cosx\left)\right(3secxtanx+4secxcotx\left)\right]$
$=-6+15tanx+8cotx+206ta{n}^{2}x-8cotx-15tanx+20co{t}^{2}x$
$=-6+20-6(se{c}^{2}x-1)+2O(cose{c}^{2}x-1)$
$=-6+20-6se{c}^{2}x+6+20cose{c}^{2}x-20$
$=-6se{c}^{2}x+20cose{c}^{2}x$
2nd method
$=\frac{d}{dx}\left[\right(3secx-4cosecx\left)\right(-2sinx+5cosx\left)\right]$
$=\frac{d}{dx}(-6secxsinx+15secxcosx+8cosecxsinx-20cosecxcosx)$
$=\frac{d}{dx}(-6\frac{sinx}{cosx}+15\frac{cosx}{cosx}+8\frac{sinx}{sinx}-20\frac{cosx}{sinx})$
$=\frac{d}{dx}(-6tanx-20cotx+23)$
$=-6se{c}^{2}x+20cose{c}^{2}x$
Using both the methods, we get the same answer.