What Are Higher Order Derivatives? So far we’ve only dealt with first derivatives . But we can keep taking derivatives of derivatives, which gives us higher order derivatives .
First derivative f ′ ( x ) f'(x) f ′ ( x ) Second derivative f ′ ′ ( x ) f''(x) f ′′ ( x ) Third derivative f ′ ′ ′ ( x ) f'''(x) f ′′′ ( x ) These are essential for analyzing function behavior in depth, especially the first and second.
Notation Prime notation
f ′ ( x ) = first derivative f'(x) = \text{first derivative} f ′ ( x ) = first derivative f ′ ′ ( x ) = second derivative f''(x) = \text{second derivative} f ′′ ( x ) = second derivative f ′ ′ ′ ( x ) = third derivative f'''(x) = \text{third derivative} f ′′′ ( x ) = third derivative f ( n ) ( x ) = nth derivative f^{(n)}(x) = \text{nth derivative} f ( n ) ( x ) = nth derivative For higher orders (like the 12th derivative), writing twelve prime marks is ridiculous, so we use
f ( 12 ) ( x ) f^{(12)}(x) f ( 12 ) ( x ) .
Leibniz notation
d y d x = first derivative \frac{dy}{dx} = \text{first derivative} d x d y = first derivative d 2 y d x 2 = second derivative \frac{d^2y}{dx^2} = \text{second derivative} d x 2 d 2 y = second derivative d 3 y d x 3 = third derivative \frac{d^3y}{dx^3} = \text{third derivative} d x 3 d 3 y = third derivative d n y d x n = nth derivative \frac{d^n y}{dx^n} = \text{nth derivative} d x n d n y = nth derivative When a function is written as
y = f ( x ) y = f(x) y = f ( x ) , you can put the primes right on
y y y :
y ′ y ′ ′ y ′ ′ ′ y ( n ) y' \qquad y'' \qquad y''' \qquad y^{(n)} y ′ y ′′ y ′′′ y ( n ) So
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y ,
f ′ ′ ( x ) f''(x) f ′′ ( x ) , and
y ′ ′ y'' y ′′ all mean the same thing: the second derivative. Use whichever notation the problem uses.
Higher Order Derivatives of Polynomials The process is simple: take the derivative of the previous derivative.
Example 1
Find the first four derivatives of
f ( x ) = 2 x 4 − 3 x 3 + 5 x 2 − 7 x + 4 f(x) = 2x^4 - 3x^3 + 5x^2 - 7x + 4 f ( x ) = 2 x 4 − 3 x 3 + 5 x 2 − 7 x + 4 .
f ′ ( x ) = 8 x 3 − 9 x 2 + 10 x − 7 f'(x) = 8x^3 - 9x^2 + 10x - 7 f ′ ( x ) = 8 x 3 − 9 x 2 + 10 x − 7 f ′ ′ ( x ) = 24 x 2 − 18 x + 10 f''(x) = 24x^2 - 18x + 10 f ′′ ( x ) = 24 x 2 − 18 x + 10 f ′ ′ ′ ( x ) = 48 x − 18 f'''(x) = 48x - 18 f ′′′ ( x ) = 48 x − 18 f ( 4 ) ( x ) = 48 f^{(4)}(x) = 48 f ( 4 ) ( x ) = 48 f ( 5 ) ( x ) = 0 f^{(5)}(x) = 0 f ( 5 ) ( x ) = 0 Important pattern: for polynomials, if you take enough successive derivatives you’ll eventually hit zero. That’s because polynomials have finite degree, and each derivative drops the degree by 1 until you’re left with a constant, whose derivative is zero.
Example 2
Find
g ′ ′ ′ ( x ) g'''(x) g ′′′ ( x ) for
g ( x ) = x 5 + 4 x 3 − 6 x + 9 g(x) = x^5 + 4x^3 - 6x + 9 g ( x ) = x 5 + 4 x 3 − 6 x + 9 .
g ′ ( x ) = 5 x 4 + 12 x 2 − 6 g'(x) = 5x^4 + 12x^2 - 6 g ′ ( x ) = 5 x 4 + 12 x 2 − 6 g ′ ′ ( x ) = 20 x 3 + 24 x g''(x) = 20x^3 + 24x g ′′ ( x ) = 20 x 3 + 24 x g ′ ′ ′ ( x ) = 60 x 2 + 24 g'''(x) = 60x^2 + 24 g ′′′ ( x ) = 60 x 2 + 24 Fractional and Negative Exponents For non-polynomial functions, rewriting with exponent rules makes life much easier.
Example 3: radicals and fractions
Find the second derivative of
h ( x ) = x + 1 x 3 h(x) = \sqrt{x} + \frac{1}{x^3} h ( x ) = x + x 3 1 .
Rewrite using exponents:
h ( x ) = x 1 / 2 + x − 3 h(x) = x^{1/2} + x^{-3} h ( x ) = x 1/2 + x − 3 First derivative:
h ′ ( x ) = 1 2 x − 1 / 2 − 3 x − 4 h'(x) = \frac{1}{2}x^{-1/2} - 3x^{-4} h ′ ( x ) = 2 1 x − 1/2 − 3 x − 4 Second derivative, applying the power rule again:
h ′ ′ ( x ) = 1 2 ⋅ ( − 1 2 ) x − 3 / 2 − 3 ( − 4 ) x − 5 = − 1 4 x − 3 / 2 + 12 x − 5 h''(x) = \frac{1}{2} \cdot \left(-\frac{1}{2}\right)x^{-3/2} - 3(-4)x^{-5} = -\frac{1}{4}x^{-3/2} + 12x^{-5} h ′′ ( x ) = 2 1 ⋅ ( − 2 1 ) x − 3/2 − 3 ( − 4 ) x − 5 = − 4 1 x − 3/2 + 12 x − 5 Rewriting in standard form (optional):
h ′ ′ ( x ) = − 1 4 x 3 / 2 + 12 x 5 = − 1 4 x x + 12 x 5 h''(x) = -\frac{1}{4x^{3/2}} + \frac{12}{x^5} = -\frac{1}{4x\sqrt{x}} + \frac{12}{x^5} h ′′ ( x ) = − 4 x 3/2 1 + x 5 12 = − 4 x x 1 + x 5 12 Note: keep things in exponent form while you’re taking successive derivatives so you don’t drown in fraction manipulation. Convert to standard form only at the very end if you need to.
Example 4: mixed function
Find
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y for
y = 3 x − 1 + 2 x 1 / 3 y = 3x^{-1} + 2x^{1/3} y = 3 x − 1 + 2 x 1/3 .
d y d x = − 3 x − 2 + 2 3 x − 2 / 3 \frac{dy}{dx} = -3x^{-2} + \frac{2}{3}x^{-2/3} d x d y = − 3 x − 2 + 3 2 x − 2/3 d 2 y d x 2 = 6 x − 3 − 4 9 x − 5 / 3 = 6 x 3 − 4 9 x 5 / 3 \frac{d^2y}{dx^2} = 6x^{-3} - \frac{4}{9}x^{-5/3} = \frac{6}{x^3} - \frac{4}{9x^{5/3}} d x 2 d 2 y = 6 x − 3 − 9 4 x − 5/3 = x 3 6 − 9 x 5/3 4 Heads up: each higher order derivative only exists where the one before it can be differentiated. Take
f ( x ) = x 4 / 3 f(x) = x^{4/3} f ( x ) = x 4/3 :
f ′ ( x ) = 4 3 x 1 / 3 f ′ ′ ( x ) = 4 9 x − 2 / 3 = 4 9 x 2 3 f'(x) = \frac{4}{3}x^{1/3} \qquad f''(x) = \frac{4}{9}x^{-2/3} = \frac{4}{9\sqrt[3]{x^2}} f ′ ( x ) = 3 4 x 1/3 f ′′ ( x ) = 9 4 x − 2/3 = 9 3 x 2 4 f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0 exists, but
f ′ ′ ( 0 ) f''(0) f ′′ ( 0 ) would mean dividing by zero. So
f f f has a first derivative at
x = 0 x = 0 x = 0 but no second derivative there.
Second Derivative with Implicit Differentiation This one is trickier, because the first derivative usually contains
d y d x \frac{dy}{dx} d x d y terms that you have to
substitute when finding the second derivative.
The process
Find d y d x \frac{dy}{dx} d x d y To get d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y d y d x \frac{dy}{dx} d x d y When d y d x \frac{dy}{dx} d x d y Simplify Example 5
Find
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y for
sin ( y ) = x + y \sin(y) = x + y sin ( y ) = x + y .
Step 1: first derivative. Differentiate both sides with respect to
x x x :
cos ( y ) ⋅ d y d x = 1 + d y d x \cos(y) \cdot \frac{dy}{dx} = 1 + \frac{dy}{dx} cos ( y ) ⋅ d x d y = 1 + d x d y Collect
d y d x \frac{dy}{dx} d x d y terms:
cos ( y ) ⋅ d y d x − d y d x = 1 \cos(y) \cdot \frac{dy}{dx} - \frac{dy}{dx} = 1 cos ( y ) ⋅ d x d y − d x d y = 1 Factor:
d y d x [ cos ( y ) − 1 ] = 1 \frac{dy}{dx}[\cos(y) - 1] = 1 d x d y [ cos ( y ) − 1 ] = 1 Solve:
d y d x = 1 cos ( y ) − 1 \frac{dy}{dx} = \frac{1}{\cos(y) - 1} d x d y = cos ( y ) − 1 1 Step 2: second derivative. Rewrite as:
d y d x = [ cos ( y ) − 1 ] − 1 \frac{dy}{dx} = [\cos(y) - 1]^{-1} d x d y = [ cos ( y ) − 1 ] − 1 Differentiate with the chain rule:
d 2 y d x 2 = − 1 [ cos ( y ) − 1 ] − 2 ⋅ d d x [ cos ( y ) − 1 ] \frac{d^2y}{dx^2} = -1[\cos(y) - 1]^{-2} \cdot \frac{d}{dx}[\cos(y) - 1] d x 2 d 2 y = − 1 [ cos ( y ) − 1 ] − 2 ⋅ d x d [ cos ( y ) − 1 ] The derivative of
cos ( y ) − 1 \cos(y) - 1 cos ( y ) − 1 needs the chain rule too:
d d x [ cos ( y ) − 1 ] = − sin ( y ) ⋅ d y d x \frac{d}{dx}[\cos(y) - 1] = -\sin(y) \cdot \frac{dy}{dx} d x d [ cos ( y ) − 1 ] = − sin ( y ) ⋅ d x d y So:
d 2 y d x 2 = − [ cos ( y ) − 1 ] − 2 ⋅ ( − sin ( y ) ⋅ d y d x ) = sin ( y ) [ cos ( y ) − 1 ] 2 ⋅ d y d x \frac{d^2y}{dx^2} = -[\cos(y) - 1]^{-2} \cdot \left(-\sin(y) \cdot \frac{dy}{dx}\right) = \frac{\sin(y)}{[\cos(y) - 1]^2} \cdot \frac{dy}{dx} d x 2 d 2 y = − [ cos ( y ) − 1 ] − 2 ⋅ ( − sin ( y ) ⋅ d x d y ) = [ cos ( y ) − 1 ] 2 sin ( y ) ⋅ d x d y Step 3: substitute the first derivative. Replace
d y d x \frac{dy}{dx} d x d y with
1 cos ( y ) − 1 \frac{1}{\cos(y) - 1} c o s ( y ) − 1 1 :
d 2 y d x 2 = sin ( y ) [ cos ( y ) − 1 ] 2 ⋅ 1 cos ( y ) − 1 \frac{d^2y}{dx^2} = \frac{\sin(y)}{[\cos(y) - 1]^2} \cdot \frac{1}{\cos(y) - 1} d x 2 d 2 y = [ cos ( y ) − 1 ] 2 sin ( y ) ⋅ cos ( y ) − 1 1 Step 4: simplify
d 2 y d x 2 = sin ( y ) [ cos ( y ) − 1 ] 3 \frac{d^2y}{dx^2} = \frac{\sin(y)}{[\cos(y) - 1]^3} d x 2 d 2 y = [ cos ( y ) − 1 ] 3 sin ( y ) Example 6
Find
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y for
x 2 + y 2 = 16 x^2 + y^2 = 16 x 2 + y 2 = 16 .
Step 1: first derivative
2 x + 2 y d y d x = 0 ⇒ d y d x = − x y 2x + 2y\frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y} 2 x + 2 y d x d y = 0 ⇒ d x d y = − y x Step 2: second derivative. Use the quotient rule on
− x y -\frac{x}{y} − y x :
d 2 y d x 2 = − ( 1 ) ( y ) − ( x ) ( d y d x ) y 2 \frac{d^2y}{dx^2} = -\frac{(1)(y) - (x)\left(\frac{dy}{dx}\right)}{y^2} d x 2 d 2 y = − y 2 ( 1 ) ( y ) − ( x ) ( d x d y ) Step 3: substitute d y d x = − x y \frac{dy}{dx} = -\frac{x}{y} d x d y = − y x d 2 y d x 2 = − y − x ( − x y ) y 2 = − y + x 2 y y 2 \frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} d x 2 d 2 y = − y 2 y − x ( − y x ) = − y 2 y + y x 2 Step 4: simplify. Multiply the numerator by
y y \frac{y}{y} y y :
= − y 2 + x 2 y 3 = -\frac{y^2 + x^2}{y^3} = − y 3 y 2 + x 2 Since
x 2 + y 2 = 16 x^2 + y^2 = 16 x 2 + y 2 = 16 :
d 2 y d x 2 = − 16 y 3 \frac{d^2y}{dx^2} = -\frac{16}{y^3} d x 2 d 2 y = − y 3 16 Key Pattern for Implicit Second Derivatives You’ll always end up with d y d x \frac{dy}{dx} d x d y Substitute the expression you found for d y d x \frac{dy}{dx} d x d y Simplify, often using the original equation Higher Order Derivatives of Exponential and Trig Functions Example 7
Find
f ′ ′ ( x ) f''(x) f ′′ ( x ) for
f ( x ) = e x 2 f(x) = e^{x^2} f ( x ) = e x 2 .
Step 1: first derivative. Chain rule:
f ′ ( x ) = 2 x e x 2 f'(x) = 2xe^{x^2} f ′ ( x ) = 2 x e x 2 Step 2: second derivative. Now
f ′ ( x ) f'(x) f ′ ( x ) is a product, so use the product rule (and the chain rule again on
e x 2 e^{x^2} e x 2 ):
f ′ ′ ( x ) = 2 ⋅ e x 2 + 2 x ⋅ 2 x e x 2 = ( 2 + 4 x 2 ) e x 2 f''(x) = 2 \cdot e^{x^2} + 2x \cdot 2xe^{x^2} = (2 + 4x^2)e^{x^2} f ′′ ( x ) = 2 ⋅ e x 2 + 2 x ⋅ 2 x e x 2 = ( 2 + 4 x 2 ) e x 2 Common mistake: treating
2 x e x 2 2xe^{x^2} 2 x e x 2 like it’s just
e x 2 e^{x^2} e x 2 and skipping the product rule. Once the chain rule puts a factor out front, the next derivative almost always needs the product rule.
Example 8
Find
d 27 d x 27 [ sin x ] \frac{d^{27}}{dx^{27}}[\sin x] d x 27 d 27 [ sin x ] .
Derivatives of sine repeat every four steps:
d d x [ sin x ] = cos x d 2 d x 2 [ sin x ] = − sin x d 3 d x 3 [ sin x ] = − cos x d 4 d x 4 [ sin x ] = sin x \frac{d}{dx}[\sin x] = \cos x \qquad \frac{d^2}{dx^2}[\sin x] = -\sin x \qquad \frac{d^3}{dx^3}[\sin x] = -\cos x \qquad \frac{d^4}{dx^4}[\sin x] = \sin x d x d [ sin x ] = cos x d x 2 d 2 [ sin x ] = − sin x d x 3 d 3 [ sin x ] = − cos x d x 4 d 4 [ sin x ] = sin x Since
27 = 4 ( 6 ) + 3 27 = 4(6) + 3 27 = 4 ( 6 ) + 3 , the 27th derivative lands in the same spot as the 3rd:
d 27 d x 27 [ sin x ] = − cos x \frac{d^{27}}{dx^{27}}[\sin x] = -\cos x d x 27 d 27 [ sin x ] = − cos x Example 9
Find a formula for
y ( n ) y^{(n)} y ( n ) if
y = e 2 x y = e^{2x} y = e 2 x .
Each derivative brings down another factor of 2 from the chain rule:
y ′ = 2 e 2 x y ′ ′ = 4 e 2 x y ′ ′ ′ = 8 e 2 x y ( n ) = 2 n e 2 x y' = 2e^{2x} \qquad y'' = 4e^{2x} \qquad y''' = 8e^{2x} \qquad y^{(n)} = 2^n e^{2x} y ′ = 2 e 2 x y ′′ = 4 e 2 x y ′′′ = 8 e 2 x y ( n ) = 2 n e 2 x Key Takeaways Higher order derivatives: just keep taking derivatives of derivativesNotation: f ′ ′ ( x ) f''(x) f ′′ ( x ) d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y f ( n ) ( x ) f^{(n)}(x) f ( n ) ( x ) d n y d x n \frac{d^n y}{dx^n} d x n d n y More notation: for y = f ( x ) y = f(x) y = f ( x ) y ′ ′ y'' y ′′ y ( n ) y^{(n)} y ( n ) Polynomial pattern: successive derivatives of polynomials eventually hit zeroOther patterns: derivatives of sin x \sin x sin x cos x \cos x cos x e k x e^{kx} e k x k k k Existence: f ′ ′ f'' f ′′ f ′ f' f ′ x 4 / 3 x^{4/3} x 4/3 x = 0 x = 0 x = 0 Keep exponent form: while taking multiple derivatives, leave things as x n x^n x n Implicit differentiation requires substitution: when finding d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y d y d x \frac{dy}{dx} d x d y Chain rule applies: differentiating anything with y y y d y d x \frac{dy}{dx} d x d y The original equation helps: use it to simplify your final answer when you can
