Calculus: The Language of Change

Exercise Set 7.2
Implicit Differentiation

by John Robeson

Exercise 1) Implicit Drill

Verify the following implicit total differential calculations:

(a)

3 (x^2) + ((y^2)/5) = 1

Solution to (a)

Clear[x, y, dEq, dx, dy];
dEq = Dt[ 3 (x^2) + ((y^2)/5) == 1, x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

(b)

xy = 1

Solution to (b)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x y == 1, x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

(c)

x + x y = 2 y

Solution to (c)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x  +  x y  ==  2 y, x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

(d)

y + (Sqrt[y]) = x + (x^2)

Solution to (d)

Clear[x, y, dEq, dx, dy];
dEq = Dt[y + (Sqrt[y]) == x + (x^2), x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

(e)

(e^x) = Log[y]

Solution to (e)

Clear[x, y, dEq, dx, dy];
dEq = Dt[(E^x)  ==  Log[y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

(f)

x = Sin[x y]

Solution to (f)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x  ==  Sin[x y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

(g)

x = Log[Cos[3 x + 5 y]]

Solution to (g)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x  ==  Log[Cos[3 x  +  5 y]], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])

Exercise 2) More Implicit Drill

Find the total differential and solve for (dy/dx)

(a)

(x^2) - y^2 = 3

Solution to (a)

Clear[x, y, dEq, dx, dy];
dEq = Dt[(x^2) - y^2 == 3, x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(b)

y + (Sqrt[y]) = (1/x)

Solution to (b)

Clear[x, y, dEq, dx, dy];
dEq = Dt[y + (Sqrt[y]) == (1/x), x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(c)

x y = 4

Solution to (c)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x y  ==  4, x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(d)

y = Cos[x y]

Solution to (d)

Clear[x, y, dEq, dx, dy];
dEq = Dt[y  ==  Cos[x y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(e)

Sin[x] Cos[y] = (1/2)

Solution to (e)

Clear[x, y, dEq, dx, dy];
dEq = Dt[Sin[x] Cos[y]  ==  (1/2), x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(f)

y = Sin[x + y]

Solution to (f)

Clear[x, y, dEq, dx, dy];
dEq = Dt[y  ==  Sin[x  +  y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(g)

y = (e^(x y))

Solution to (g)

Clear[x, y, dEq, dx, dy];
dEq = Dt[y == (E^(x y)), x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(h)

(e^x) (e^y) = 1

Solution to (h)

Clear[x, y, dEq, dx, dy];
dEq = Dt[(E^x) (E^y) == 1, x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(i)

(e^x) = x + (y^2)

Solution to (i)

Clear[x, y, dEq, dx, dy];
dEq = Dt[(E^x) == x + (y^2), x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(j)

x = Log[x y]

Solution to (j)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x  ==  Log[x y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(k)

y = Log[(x^2) y]

Solution to (k)

Clear[x, y, dEq, dx, dy];
dEq = Dt[y == Log[(x^2) y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]

(l)

x = Log[x + y]

Solution to (l)

Clear[x, y, dEq, dx, dy];
dEq = Dt[x  ==  Log[x  +  y], x]/.Dt[y, x]->(Underscript[\(dy\+_\), dx])Solve[dEq,(Underscript[\(dy\+_\), dx])]