# How do you implicitly differentiate -1=xy^2+x^2y-e^ysec(xy) ?

May 13, 2017

$y ' = \frac{{y}^{2} + 2 x y - {e}^{y} y \sec \left(x y\right) \tan \left(x y\right)}{{e}^{y} x \sec \left(x y\right) \tan \left(x y\right) + {e}^{y} \sec \left(x y\right) - 2 x y - {y}^{2}}$

#### Explanation:

Chain Rule:
if $f \left(x\right) = g \left(h \left(x\right)\right)$ then $f ' \left(x\right) = g ' \left(h \left(x\right)\right) h ' \left(x\right)$
Product Rule:
if $\left[f \left(x\right) g \left(x\right)\right] ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$
The fact that:
$\left[\frac{d}{\mathrm{dx}}\right] {e}^{x} = {e}^{x}$
And $\left[\frac{d}{\mathrm{dx}}\right] \sec \left(x\right) = \sec \left(x\right) \tan \left(x\right)$

Solving:
$\frac{d}{\mathrm{dx}} \left[- 1\right] = \frac{d}{\mathrm{dx}} \left[x {y}^{2} + {x}^{2} y - {e}^{y} \sec \left(x y\right)\right]$

It helps to split:
(1)
$\frac{d}{\mathrm{dx}} \left[x {y}^{2}\right] = \left(x\right) \left(\frac{d}{\mathrm{dx}} \left[{y}^{2}\right]\right) + \left(1\right) \left({y}^{2}\right) = 2 x y y ' + {y}^{2}$

(2)
$\frac{d}{\mathrm{dx}} \left[{x}^{2} y\right] = \left({x}^{2}\right) \left(\frac{d}{\mathrm{dx}} \left[y\right]\right) + \left(2 x\right) \left(y\right) = {x}^{2} \left(y '\right) \left(1\right) + 2 x y = {x}^{2} y ' + 2 x y$

(3a) Before proceeding, it is better to solve $\frac{d}{\mathrm{dx}} \left[\sec \left(x y\right)\right]$
Since you need to product rule the 3rd term, then chain rule $\sec \left(x y\right)$ then product rule again $x y$ which can get messy.

$\frac{d}{\mathrm{dx}} \left[\sec \left(x y\right)\right] = \left(\sec \left(x y\right) \tan \left(x y\right)\right) \left(\frac{d}{\mathrm{dx}} \left[x y\right]\right) = \sec \left(x y\right) \tan \left(x y\right) \left(x y ' + y\right)$

(3b)
d/dx[e^y sec(xy)]=e^y(d/dx[sec(xy)])+(d/dx[e^y])(sec(xy))=e^y(sec(xy)tan(xy)(xy'+y))+(e^yy')(sec(xy))=e^yxy'sec(xy)tan(xy)+ e^yysec(xy)tan(xy)+(e^yy')sec(xy)

Now putting it all together
$\frac{d}{\mathrm{dx}} \left[- 1\right] = \frac{d}{\mathrm{dx}} \left[x {y}^{2} + {x}^{2} y - {e}^{y} \sec \left(x y\right)\right]$
0=[2xyy'+y^2]+[x^2y'+2xy]-[e^yxy'sec(xy)tan(xy)+ e^yysec(xy)tan(xy)+(e^yy')sec(xy)]
Put all terms with $y '$ on one side and those without in the other.
0=[2xyy'+y^2]+[x^2y'+2xy]-[e^yxy'sec(xy)tan(xy)+ e^yysec(xy)tan(xy)+e^yy'sec(xy)]
$- 2 x y y ' - {x}^{2} y ' + {e}^{y} x y ' \sec \left(x y\right) \tan \left(x y\right) + {e}^{y} y ' \sec \left(x y\right) = {y}^{2} + 2 x y - {e}^{y} y \sec \left(x y\right) \tan \left(x y\right)$
$y ' \left(- 2 x y - {x}^{2} + {e}^{y} x \sec \left(x y\right) \tan \left(x y\right) + {e}^{y} \sec \left(x y\right)\right) = {y}^{2} + 2 x y - {e}^{y} y \sec \left(x y\right) \tan \left(x y\right)$
$y ' = \frac{{y}^{2} + 2 x y - {e}^{y} y \sec \left(x y\right) \tan \left(x y\right)}{{e}^{y} x \sec \left(x y\right) \tan \left(x y\right) + {e}^{y} \sec \left(x y\right) - 2 x y - {x}^{2}}$

Tips: It helps to break down terms and solve each. Be mindful of parantheses And watch out for rules to apply especially if there functions within a function.