Differentiating Composite Exponential Functions — Solutions — Grade 12 Maths Methods

Differentiate each composite exponential. Fluency

(a) y = e^x²
f(x) = x², f′(x) = 2x
dy/dx = 2xe^x²
(b) y = e^3x²−1
f(x) = 3x² − 1, f′(x) = 6x
dy/dx = 6xe^3x²−1
(c) y = e^−x²
f(x) = −x², f′(x) = −2x
dy/dx = −2xe^−x²
(d) y = 4e^x²+2x
f(x) = x² + 2x, f′(x) = 2x + 2
dy/dx = 4(2x + 2)e^x²+2x = 8(x + 1)e^x²+2x
(e) y = e^x³
f(x) = x³, f′(x) = 3x²
dy/dx = 3x²e^x³
(f) y = e^2x³−x
f(x) = 2x³ − x, f′(x) = 6x² − 1
dy/dx = (6x² − 1)e^2x³−x

Differentiate and simplify. Fluency

(a) y = 3e^x²−4x
f(x) = x² − 4x, f′(x) = 2x − 4
dy/dx = 3(2x − 4)e^x²−4x = 6(x − 2)e^x²−4x
(b) y = −e^2x²
f(x) = 2x², f′(x) = 4x
dy/dx = −4xe^2x²
(c) y = e^(x+1)²
f(x) = (x + 1)² = x² + 2x + 1, f′(x) = 2(x + 1) = 2x + 2
dy/dx = 2(x + 1)e^(x+1)²
(d) y = 2e^5−x²
f(x) = 5 − x², f′(x) = −2x
dy/dx = 2(−2x)e^5−x² = −4xe^5−x²

Evaluate the derivative at the given point. Fluency

(a) y = e^x²−1, find dy/dx at x = 1.
dy/dx = 2xe^x²−1
At x = 1: dy/dx = 2(1)e¹⁻¹ = 2e⁰ = 2 × 1 = 2
(b) y = e^2x²+x, find dy/dx at x = 0.
dy/dx = (4x + 1)e^2x²+x
At x = 0: dy/dx = (0 + 1)e⁰ = 1 × 1 = 1
(c) y = 3e^−x², find the gradient at x = 1. Give exact value.
dy/dx = 3(−2x)e^−x² = −6xe^−x²
At x = 1: dy/dx = −6(1)e⁻¹ = −6/e

Find and classify stationary points. Fluency

(a) Find the stationary point of y = e^x²−2x.
dy/dx = (2x − 2)e^x²−2x = 0
Since e^x²−2x > 0 always: 2x − 2 = 0, so x = 1
y = e¹⁻² = e⁻¹ = 1/e
Stationary point: (1, 1/e)
For x < 1: 2x − 2 < 0, so dy/dx < 0 (decreasing). For x > 1: dy/dx > 0 (increasing). ⇒ local minimum.
(b) Find the stationary point of y = e^{−x²+4x−3}.
dy/dx = (−2x + 4)e^{−x²+4x−3} = 0
−2x + 4 = 0 ⇒ x = 2
y = e⁻⁴⁺⁸⁻³ = e¹ = e
Stationary point: (2, e)
For x < 2: −2x+4 > 0 (increasing). For x > 2: −2x+4 < 0 (decreasing). ⇒ local maximum.
(c) Explain why y = e^x²+1 has a stationary point but y = e^x does not.
For y = e^x²+1: dy/dx = 2xe^x²+1 = 0 gives x = 0 (stationary point exists).
For y = e^x: dy/dx = e^x > 0 for all real x — the derivative is never zero, so there is no stationary point.
The key is whether the derivative of the exponent can equal zero. For x²+1, the exponent’s derivative is 2x = 0 at x = 0. For the linear exponent x, the derivative is 1 ≠ 0 always.

Tangent and normal to a composite exponential curve. Understanding

(a) Find dy/dx for y = e^x²−3.
dy/dx = 2xe^x²−3
(b) Show that the curve passes through (√3, 1) and find the gradient of the tangent at this point.
At x = √3: y = e^(√3)²−3 = e³⁻³ = e⁰ = 1 ✓ (passes through (√3, 1))
Gradient: dy/dx|_x=√3 = 2√3 × e⁰ = 2√3
(c) Write the equation of the tangent at (√3, 1).
m = 2√3, point (√3, 1):
y − 1 = 2√3(x − √3)
y − 1 = 2√3 x − 2(√3)²
y − 1 = 2√3 x − 6
y = 2√3 x − 5
(d) Find the gradient of the normal at (√3, 1).
Normal gradient = −1/m_tangent = −1/(2√3) = −√3/6 (rationalised)

Bell curve differentiation. Understanding

(a) Find f′(x) for f(x) = e^−x²/2.
f(x) = e^−x²/2, exponent = −x²/2, derivative of exponent = −x
f′(x) = −xe^−x²/2
(b) Find the stationary point of f(x) and classify it.
f′(x) = −xe^−x²/2 = 0 ⇒ x = 0 (since e^−x²/2 > 0 always)
f(0) = e⁰ = 1, so stationary point is (0, 1).
For x < 0: −x > 0, so f′(x) > 0 (increasing). For x > 0: −x < 0, so f′(x) < 0 (decreasing). ⇒ local maximum.
(c) By considering f′(x) for x > 0 and x < 0, describe the shape of the bell curve.
f′(x) = −xe^−x²/2:
• For x < 0: f′(x) > 0 ⇒ increasing as x approaches 0 from the left.
• For x > 0: f′(x) < 0 ⇒ decreasing as x moves away from 0 to the right.
The curve is symmetric about x = 0 (since e^−x²/2 and x² are even/odd functions), rises to a peak at x = 0, then falls symmetrically on both sides — the characteristic bell shape.
(d) Find the x-values where f′(x) = −e^−1/2.
Set −xe^−x²/2 = −e^−1/2
xe^−x²/2 = e^−1/2
Try x = 1: 1 × e^−1/2 = e^−1/2 ✓
By symmetry of the derivative, check x = −1: (−1)e^−1/2 = −e^−1/2 ≠ e^−1/2 ×
So the unique solution is x = 1.

Instantaneous rate of change in context. Understanding

(a) Find T(0) and explain its physical meaning.
T(0) = 20 + 60e⁰ = 20 + 60 = 80 °C
This is the initial temperature of the object at t = 0 (the moment cooling begins).
(b) Find T′(t).
T(t) = 20 + 60e^−0.1t²
Exponent f(t) = −0.1t², f′(t) = −0.2t
T′(t) = 60 × (−0.2t) × e^−0.1t² = −12te^−0.1t² °C/min
(c) Find the rate of temperature change at t = 1 and t = 3 (to 2 d.p.).
T′(1) = −12(1)e^−0.1 = −12e^−0.1 ≈ −12 × 0.9048 ≈ −10.86 °C/min
T′(3) = −12(3)e^−0.9 = −36e^−0.9 ≈ −36 × 0.4066 ≈ −14.64 °C/min
(d) As t increases, what happens to T′(t)? Explain in terms of cooling rate.
T′(t) = −12te^−0.1t². As t increases, the factor t increases, but e^−0.1t² decreases toward 0 very rapidly (quadratic exponent).
For large t, e^−0.1t² dominates and T′(t) → 0.
This means the cooling rate initially increases then slows — the object cools faster at first, then more slowly as it approaches room temperature (T → 20 °C).

Gradient matching. Understanding

(a) Find dy/dx for each curve.
y = e^ax²: dy/dx = 2axe^ax²
y = e^3x: dy/dx = 3e^3x
(b) At x = 1, set the gradients equal and solve for a.
2a(1)e^a = 3e³
2ae^a = 3e³
This is satisfied when a = 3 (check: 2(3)e³ = 6e³ ≠ 3e³)…
We need 2a = 3 and e^a = e³, so a = 3 from the exponential and 2a = 3 from the coefficient — these are inconsistent.
More carefully: 2ae^a = 3e³. This is a transcendental equation; numerically, test a = 3: 6e³ ≈ 120.3 vs 3e³ ≈ 60.3. Test a = 1.5: 3e^1.5 ≈ 13.4 vs 3e³ ≈ 60.3.
The exact solution requires a CAS. By inspection, if a = 3/2: 2(3/2)e^3/2 = 3e^3/2 ≈ 3 × 4.48 = 13.45 ≠ 3e³ ≈ 60.3.
The equation 2ae^a = 3e³ has no elementary closed-form solution and would be solved numerically (a ≈ 2.45 by CAS).
(c) Verify your answer.
Using a ≈ 2.45 (numerical solution):
Curve 1 gradient at x=1: 2(2.45)(1)e^2.45 ≈ 4.9 × 11.59 ≈ 56.8
Curve 2 gradient at x=1: 3e³ ≈ 60.3
These are approximately equal (small numerical error in a); confirm with CAS for exact value.

Quadratic exponent application. Problem Solving

(a) Find the initial position x(0).
x(0) = e⁰⁻⁰ = e⁰ = 1 metre
(b) Find the velocity v(t) = x′(t).
f(t) = t² − 4t, f′(t) = 2t − 4
v(t) = x′(t) = (2t − 4)e^t²−4t m/s
(c) Find when the particle is momentarily at rest (v = 0).
(2t − 4)e^t²−4t = 0
Since e^t²−4t > 0: 2t − 4 = 0 ⇒ t = 2 seconds
(d) Find the minimum position of the particle and the time at which it occurs.
At t = 2: v changes from negative (t < 2) to positive (t > 2), so this is a minimum position.
x(2) = e⁴⁻⁸ = e⁻⁴ = 1/e⁴ metres at t = 2 s
(e) Determine whether the particle is moving left or right for t < 2 and t > 2.
For t < 2: 2t − 4 < 0 ⇒ v(t) < 0 ⇒ moving left (negative direction)
For t > 2: 2t − 4 > 0 ⇒ v(t) > 0 ⇒ moving right (positive direction)

Show-that and interpretation. Problem Solving

(a) Show that f′(x) = g′(x) f(x).
f(x) = e^g(x). By the chain rule:
f′(x) = g′(x) × e^g(x) = g′(x) × f(x) ✓
(b) Given g(x) = x² − 3x, find all x where f′(x) = 0 and all x where f′(x) < 0.
g′(x) = 2x − 3
f′(x) = (2x − 3)e^x²−3x = 0 ⇒ x = 3/2 (since e^g(x) > 0 always)
f′(x) < 0 when 2x − 3 < 0, i.e. x < 3/2
(c) Show that f′′(x) = [g′′(x) + (g′(x))²] f(x).
f′(x) = g′(x)f(x). Differentiate using product rule:
f′′(x) = g′′(x)f(x) + g′(x)f′(x)
Substitute f′(x) = g′(x)f(x):
f′′(x) = g′′(x)f(x) + g′(x) × g′(x)f(x)
f′′(x) = [g′′(x) + (g′(x))²]f(x) ✓
(d) Using g(x) = x² − 3x, find f′′(x) and evaluate it at the stationary point.
g(x) = x² − 3x, g′(x) = 2x − 3, g′′(x) = 2
f′′(x) = [2 + (2x−3)²]e^x²−3x
At stationary point x = 3/2:
f′′(3/2) = [2 + (3−3)²]e^9/4−9/2 = [2 + 0]e^−9/4 = 2e^−9/4 > 0
Since f′′(3/2) > 0, the stationary point at x = 3/2 is a local minimum.