Implicit Differentiation

Selected Problems

Matthew Staley

September 20, 2011

Implicit Differentiation : Selected Problems

1. Find dy/dx.
√
3
(a) y= 2x − 5
= (2x − 5)1/3
dy 1
= (2x − 5)1/3−1 (2)
dx 3
2
= (2x − 5)−2/3
3

!
3
(b) y= 2 + tan (x2 )
= (2 + tan (x2 ))1/3
dy 1
= (2 + tan (x2 ))(1/3−1) (sec2 (x2 ))(2x)
dx 3
2
= x sec2 (x2 )(2 + tan (x2 ))−2/3
3

(c) y = x3 (5x2 + 1)−2/3

" #
dy 2 −5/3
3
= x − (5x + 1)2
(10x) + 3x2 (5x2 + 1)−2/3
dx 3
−20x4 3x2
= +
3(5x2 + 1)5/3 (5x2 + 1)2/3
−20x4 3x2 3(5x2 + 1)
= + ·
3(5x2 + 1)5/3 (5x2 + 1)2/3 3(5x2 + 1)
−20x4 9x2 (5x2 + 1)
= +
3(5x2 + 1)5/3 3(5x2 + 1)5/3
−20x4 + 45x4 + 9x2
=
3(5x2 + 1)5/3
25x4 + 9x2
=
3(5x2 + 1)5/3
1 2 2
= x (5x + 1)−5/3 (25x2 + 9)
3

1
2. Given x + xy − 2x3 = 2.

(a) Find dy/dx by differentiating implicitly.

d d
(x + xy − 2x3 ) = (2)
dx dx
d d d
(x) + (xy) − (2x3 ) = 0
dx dx dx
d
1 + x (y) + y − 6x2 = 0
dx
dy
1 + x + y − 6x2 = 0
dx

(b) Solve the equation for y as a function of x, and find dy/dx from that
equation.

x + xy − 2x3 =2
xy = 2 + 2x3 − x
y = 2x−1 + 2x2 − 1 (∗)
dy
= −2x−2 + 4x
dx

(c) Confirm that the two results are consistent by expressing the derivative
in part (a) as a function of x alone.

dy
1+x + y − 6x2 = 0
dx
dy
x = 6x2 − y − 1
dx
dy
= 6x − x−1 (y) − x−1 Sub (∗) in part (b) into y here.
dx
= 6x − x−1 (2x−1 + 2x2 − 1) − x−1
= 6x − 2x−2 − 2x + x−1 − x−1
= −2x−2 + 4x

dy 2
We see that (b) and (c) each give = 4x − 2 .
dx x

2
3. Find dy/dx by implicit differentiation.
(a) x3 + y 3 = 3xy 2

d 3 d
(x + y 3 ) = (3xy 2 )
dx dx
dy dy
3x2 + 3y 2 = 3y 2 + 3x(2y)
dx dx
dy
(3y 2 − 6xy) = 3y 2 − 3x2
dx
dy 3y 2 − 3x2 y 2 − x2
= 2 = 2
dx 3y − 6xy y − 2xy

(b) x2 y + 3xy 3 − x = 3

d 2 d
(x y + 3xy 3 − x) = (3)
dx dx
dy dy
2xy + x2 + 3y 3 + 3x(3y 2 ) − 1 = 0
dx dx
dy 2
(x + 9xy 2 ) = 1 − 2xy − 3y 3
dx
dy 1 − 2xy + 3y 3
=
dx x2 + 9xy 2

1 1
(c) √ +√ =1
x y

d $ −1/2 % d
x + y −1/2 = (1)
dx dx
1 1 dy
− x−3/2 − y −3/2 =0
2 2 dx
1 dy 1
− 3/2 = 3/2
2y dx 2x
dy y 3/2
= − 3/2
dx x

3
(d) sin (x2 y 2 ) = x

d d
(sin (x2 y 2 )) = (x)
dx dx

d 2 2
cos (x2 y 2 ) (x y ) = 1
dx

" #
2 2 2dy 2
cos (x y ) 2xy + x (2y) =1
dx

dy 1
2xy 2 + 2x2 y =
dx cos (x2 y 2 )

dy 1
2x2 y = − 2xy 2
dx cos (x2 y 2 )

dy 1 2xy 2
= 2 −
dx 2x y cos (x2 y 2 ) 2x2 y

1 y
= −
2x2 y 2 2
cos (x y ) x

1 y 2xy cos(x2 y 2 )
= − ·
2x2 y cos (x2 y 2 ) x 2xy cos(x2 y 2 )

1 − 2xy 2 cos(x2 y 2 )
=
2x2 y cos(x2 y 2 )

4
4. Find d2 y/dx2 by implicit differentiation.
(a) 2x2 − 3y 2 = 4

d d
(2x2 − 3y 2 ) = (4)
dx dx
dy
4x − 6y =0
dx
dy
−6y = −4x
dx
dy 2x
=
dx 3y

" # " #
d dy d 2x
=
dx dx dx 3y
dy
d2 y 3y(2) − 2x(3) dx
=
dx2 (3y)2
& '
6y − 6x 2x 3y
= 2
9y
& '
2
y
(3y 2 − x2 )
=
9y 2
2(3y 2 − x2 )
=
9y 3

(b) xy + y 2 = 2

d d
(xy + y 2 ) = (2)
dx dx
dy dy
y + x + 2y = 0 (∗)
dx dx
dy
(x + 2y) = −y
dx
dy y
=−
dx x + 2y

5
Now it is easier to differentiate (∗) again instead of the above expression for dy/dx . . .
" #
d dy dy d
y + x + 2y = (0)
dx dx dx dx

dy d2 y dy d2 y dy dy
+x 2 + + 2y 2 + 2 · =0
dx dx dx dx dx dx

" #2
dy d2 y dy
2 + 2 (x + 2y) + 2 =0
dx dx dx

( " #2 )
d2 y dy dy
(x + 2y) = −2 +
dx2 dx dx
( " #2 )
y y
= −2 − + −
x + 2y x + 2y
" #
y y2
= −2 − +
x + 2y (x + 2y)2

" #
y(x + 2y) + y 2
= −2 −
(x + 2y)2

2xy + 4y 2 − 2y 2
=
(x + 2y)2

d2 y 2xy + 2y 2
(x + 2y) =
dx2 (x + 2y)2

d2 y 2y(x + y)
→ 2
=
dx (x + 2y)3

6
5. Use implicit differentiation to find the slope of the tangent line to the curve at
the specified point.
& √ '
(a) x4 + y 4 = 16;
4
1, 15

d 4 d
(x + y 4 ) = (16)
dx dx
dy
4x3 + 4y 3 =0
dx
dy x3
=− 3
dx y
(1)3
=− √ ≈ −0.1312
( 4 15)3

(b) 2(x2 + y 2 )2 = 25(x2 − y 2 ); (3, 1)

d d
(2(x2 + y 2 )2 ) = (25(x2 − y 2 ))
dx" # dx
2 2 dy dy
4(x + y ) 2x + 2y = 50x − 50y
dx dx

Rather than try and solve for dy/dx now, we can just go ahead and plug
in x = 3, y = 1 and solve for dy/dx with numbers which is much easier. . .
" #
2 2 dy dy
4(3 + 1 ) 2(3) + 2(1) = 50(3) − 50(1)
dx dx
" #
dy dy
40 6 + 2 = 150 − 50
dx dx
dy dy
240 + 80 = 150 − 50
dx dx
dy
130 = −90
dx
dy 9
= −
dx 13