When y is defined implicitly by an equation F(x,y)=0:
Differentiate both sides with respect to x
Remember that dxd(y (chain rule)
4.2 Example 1: Circle
Problem: Find dxdy if x2+
Solution:
Differentiate both sides with respect to x:
2x+2ydxdy=0
Solve for dxdy:
2ydxdy=−2x⟹
4.3 Example 2: Product Within Implicit
Problem: Find dxdy if x3+
Solution:
Differentiate term by term:
3x2+(1⋅y+x⋅d
Group terms with dxdy:
xdxdy+3y2
Factor:
dxdy(x+3y2)=dxdy=−x+3y
5. Logarithmic Differentiation
5.1 When to Use
Useful for:
Functions of the form y=[f(x)]g(x)
Products/quotients with many factors
Complex exponential forms
5.2 Method
Take natural logarithm of both sides: lny=ln(f(x))
Use logarithm properties to simplify
Differentiate implicitly with respect to x
Solve for dx
5.3 Example 1: Variable Power
Problem: Differentiate y=xx for x>0
Solution:
Take logarithm:
lny=ln(xx)=xlnx
Differentiate both sides:
y1dxdy=
Multiply by y:
dxdy=y(lnx+1)=
5.4 Example 2: Complex Product/Quotient
Problem: Differentiate y=(x+3)4(x+1)
Solution:
Take logarithm:
lny=3ln(x+1)+2ln(x−2)−4
Differentiate:
y1dxdy
Multiply by y:
dxdy=
6. Parametric Differentiation
6.1 Method
If x=f(t) and y=g(t), then:
dxdy=dx
Second derivative:
dx2d
6.2 Example
Problem: Find dxdy and dx if and
Solution:
First derivatives with respect to t:
dtdx=2t,dt
First derivative:
dxdy=2t
Second derivative:
dx2d
7. Higher-Order Derivatives
7.1 Notation
f′(
7.2 Standard n-th Derivatives
dxndn(edxndndxndndxndndxndn
7.3 Leibniz Rule for Product
dxndn
7.4 Example: Higher-Order Derivative
Problem: Find dx3d3y if
Solution:
dxdy=4x3+6xdx2d2y=12xdx3d3y=24x+
8. Applications of Derivatives
8.1 Rate of Change
If y=f(x), the rate of change of y with respect to x is d.
Example: The radius of a circle increases at 2 cm/s. How fast is the area increasing when r=5 cm?
Solution:A=πr2
dtdA=2πr
8.2 Tangent and Normal Lines
For y=f(x) at point (x1,y1):
Slope of tangent:m=f′(x1)
Equation of tangent:
y−y1=f′(x
Slope of normal:m⊥=−f′(x (if )
Equation of normal:
y−y1=−f
Example: Find the tangent to y=x2 at (2,4).
Solution:
dxdy=2x⟹m=4
Tangent: y−4=4(x−2)⟹y=4x−4
8.3 Increasing and Decreasing Functions
f′(x)>0 on interval ⇒f is increasing
on interval is
8.4 Maxima and Minima (First Derivative Test)
At critical point c where f′(c)=0:
If f′ changes from + to −: local maximum
If f′ changes from to :
8.5 Maxima and Minima (Second Derivative Test)
At critical point c:
f′′(c)<0: local maximum
f′′(c)>:
8.6 Example: Optimization
Problem: Find the dimensions of a rectangle with perimeter 40 m that maximizes area.
Solution:
Let length =l, width =w.
Constraint: 2l+2w=40⇒
Area: A=l(20−l)=20l−l2
dldA=20−2l=0⟹dl2d2A=−2
So l=10 m, w=10 m. Maximum area is a square of 100 m².
9. Rolle's Theorem and Mean Value Theorem
9.1 Rolle's Theorem
If f(x) is:
Continuous on [a,b]
Differentiable on (a,b)
f(a)=f(b)
Then there exists at least one c∈(a,b) such that:
f′(c)=0
9.2 Lagrange's Mean Value Theorem (LMVT)
If f(x) is:
Continuous on [a,b]
Differentiable on (a,b)
Then there exists at least one c∈(a,b) such that:
f′(c)=b−af(b)−f
10. Solved Examples (Mixed Practice)
Example 1: Chain Rule Nested
Problem: Differentiate y=sin3(2x+1)
Solution:
Let y=u3 where u=sinv and v=2.
dxdy=3u=6sin2(2x+1)cos(2x+1)
Example 2: Implicit with Trigonometric
Problem: Find dxdy if siny+cosx=
Solution:
Differentiate:
cosy⋅dxdy−sinx=y
Group:
dxdy(cosy−x)=y+sinxdxdy=cosy−x
Example 3: Related Rates
Problem: A ladder 5 m long rests against a wall. The bottom slides away at 2 m/s. How fast is the top sliding down when the bottom is 3 m from the wall?
Solution:
Let x = distance from wall, y = height on wall.
x2+y2=25
When x=3: y=4.
Differentiate with respect to t:
2xdtdx+2ydtd2(3)(2)+2(4)dtdy=012+8dtdy=0⟹
The negative indicates the top is sliding down at 1.5 m/s.
11. Practice Problems
Level 1 — Basic Differentiation
dxd(7x5−3x
Level 2 — Product, Quotient, Chain
dxd(x2ex)
Level 3 — Implicit, Logarithmic, Parametric
Find dxdy if x2+
Level 4 — Applications
Find the equation of the tangent to y=x3−2x at x=2
Find local maxima/minima of
12. Answers to Practice Problems
Problem
Answer
1
35x4−6x+2
2
−
Quick Reference Formula Sheet
Function
Derivative
xn
nxn−1
e
Rule
Formula
Product
(uv)′=u′v+uv
End of Differentiation Formulas Guide
n−1
f′
(
x
)
f′(x)±
g′(x)
x
4−1
=
20x3
dv
+
v⋅
dxdu
0
2
v⋅dxdu−u⋅dxdv
v2u′v−uv′
⋅
dxdu
′
(
g
(
x
))
⋅
g′(x)
cos
x
+
sinx⋅
2x=
x2cosx+
2xsinx
dy
=
(x2+1)2(x2+1)(3)−(3x+1)(2x)=
(x2+1)23x2+3−6x2−2x=
(x2+1)2−3x2−2x+3
dxdu=
6x
4
⋅
6x=
5(3x2+
1)4⋅
6x=
30x(3x2+
1)4
−
1
)
=
2x1
−x21
1
)
=
−xn+1n
a
xlna1
ax
x
+
b
a
1−x21
1−x2−1
1+x2
1
1+x2−1
xx2−11
xx2−1−1
b
))
=
1−(ax+b)2a
b
))
=
1+(ax+b)2a
x2+11
x2−11
1−x21
(
∣
x
∣
<
1)
⋅
2x=
2xex2
x2
⋅
x1+
lnx⋅
2xex2=
xex2+
2xex2lnx
2
ln
x
)
n
)
=
nyn−1dxdy
Solve for dxdy
y
2
=
25
dxdy
=
−yx
x
y
+
y3=
7
x
dy
)
+
3y2dxdy=
0
dx
dy
=
−3x2−
y
−
(
3
x2
+
y)
2
3x2+y
dy
1⋅
lnx+
x⋅
x1=
lnx+
1
xx
(
ln
x
+
1)
3
(
x
−
2
)2
ln
(
x
+
3)
=
x+13+
x−22−
x+34
(x+3)4(x+1)3(x−2)2
[x+13+x−22−x+34]
/
d
t
dy/dt
=
f′(t)g′(t)
2
y
=
dxd(dxdy)=
dtdxdtd(dxdy)
2
d2y
x=t2
y=t3−3t
dy
=
3t2−
3
3t2−3
=
2t3(t2−1)
dtd
(dxdy)
=
dtd(2t3t2−3)=
4t26t⋅2t−(3t2−3)⋅2=
4t212t2−6t2+6=
4t26t2+6=
2t23(t2+1)
2
y
=
2t23(t2+1)⋅
2t1=
4t33(t2+1)
x
)
=
dxdy,f′′(x)=
dx2d2y,f′′′(x)=
dx3d3y,f(n)(x)=
dxndny
ax
)
=
aneax
(
sin
(
a
x
+
b))=
ansin(ax+b+2nπ)
(
cos
(
a
x
+
b))=
ancos(ax+b+2nπ)
(
ln
(
x
))
=
xn(−1)n−1(n−1)!
(ax+b1)
=
(ax+b)n+1(−1)nn!an
(
uv
)
=
r=0∑n(rn)u(n−r)v(r)
y=x4+2x3−x+5
2
−
1
2
+
12x
12
x
dy
d
t
dr
=
2π(5)(2)=
20π cm2/s
1
)
(
x
−
x1)
1
)
1
f′(x1)=0
′
(
x1
)
1
(
x
−
x1)
f′(x)<
0
⇒
f
decreasing
f′(x)=0⇒critical point (possible extremum)
−
+
local minimum
If f′ does not change sign: neither (point of inflection)
0
local minimum
f′′(c)=0: test fails, use first derivative test
l
+
w=
20⇒
w=
20−
l
l
=
10
<
0⟹
maximum
(
a
)
x
+
1
2
⋅
cosv⋅
2=
3sin2(2x+
1)⋅
cos(2x+
1)⋅
2
x
y
+
xdxdy
y+sinx
y
=
0
dtdy
=
−812=
−1.5 m/s
2
+
2x−
1)
dxd(x32)
dxd(4ex+3lnx)
dxd(sinx+cosx+tanx)
dxd(x+x1)
dxd(x+1x2+1)
dxd((2x3−1)7)
dxd(ln(sinx))
dxd(xsin−1x)
x
y
+
y2=
7
Differentiate y=xsinx
Find dxdy if x=t2+1, y=t3−t
Differentiate y=(cosx)x
Find dxdy if exy=x+y
f(x)=x3−3x2+4
A spherical balloon is inflated at 10 cm³/s. Find dtdr when r=5 cm
Find dx2d2y if y=e2xsin3x
Verify Rolle's theorem for f(x)=x2−4x+3 on [1,3]
x4
6
3
4ex+x3
4
cosx−sinx+sec2x
5
2x1−2x3/21
6
ex(x2+2x)
7
(x+1)2x2+2x−1
8
42x2(2x3−1)6
9
cotx
10
sin−1x+1−x2x
11
−x+2y2x+y
12
xsinx(xsinx+cosxlnx)
13
2t3t2−1
14
(cosx)x(ln(cosx)−xtanx)
15
xexy−11−yexy or x(x+y)−11−y(x+y)
16
y=10x−16
17
Max at x=0 (y=4); Min at x=2 (y=0)
18
10π1 cm/s
19
e2x(12cos3x−5sin3x)
20
f(1)=f(3)=0; f′(c)=0 at c=2∈(1,3) ✓
x
ex
ax
axlna
lnx
x1
sinx
cosx
cosx
−sinx
tanx
sec2x
secx
secxtanx
sin−1x
1−x21
tan−1x
1+x21
sinhx
coshx
coshx
sinhx
′
Quotient
(vu)′=v2u′v−uv′
Chain
dxdf(g(x))=f′(g(x))⋅g′(x)
Logarithmic
dxdlny=y1dxdy
Parametric
dxdy=dx/dtdy/dt
Differentiation Formulas Guide: Basic to Intermediate with Examples