A-Level Mathematics — Hard Differentiation Questions
Exam-focused questions · Clear step-by-step solutions · Human-style notes
Instructions: These are challenging A-Level differentiation problems with fully worked, step-by-step
solutions. The style is written to look like human-made notes — clear headings, reasoning, and final
answer boxed.
Problem 1 — Differentiate (logarithmic + product rules)
Differentiate the function y = x^x · sin(x) with respect to x. Give a simplified form for
dy/dx.
Solution — Step 1: Use logarithmic differentiation for x^x
We have y = x^x · sin x. Treat u(x) = x^x and v(x) = sin x so y = u·v. First find u'(x) using
logarithmic differentiation: let u = x^x.
Take ln u = ln(x^x) = x ln x. Differentiate both sides w.r.t x: (1/u)·u' = 1·ln x + x·(1/x) = ln x +
1. So u' = u(ln x + 1) = x^x(ln x + 1).
Solution — Step 2: Use product rule for y = u·v
y' = u'·v + u·v'. Substitute u, u', v, v'. We computed u' = x^x(ln x + 1), u = x^x, v = sin x, v' =
cos x.
Therefore: y' = x^x(ln x + 1)·sin x + x^x·cos x Factor x^x: y' = x^x[ (ln x + 1)·sin x + cos x ].
This is the simplified derivative.
Made for Stuvia-style upload — keep previews light and watermarked. Page 1
Exam-focused questions · Clear step-by-step solutions · Human-style notes
Instructions: These are challenging A-Level differentiation problems with fully worked, step-by-step
solutions. The style is written to look like human-made notes — clear headings, reasoning, and final
answer boxed.
Problem 1 — Differentiate (logarithmic + product rules)
Differentiate the function y = x^x · sin(x) with respect to x. Give a simplified form for
dy/dx.
Solution — Step 1: Use logarithmic differentiation for x^x
We have y = x^x · sin x. Treat u(x) = x^x and v(x) = sin x so y = u·v. First find u'(x) using
logarithmic differentiation: let u = x^x.
Take ln u = ln(x^x) = x ln x. Differentiate both sides w.r.t x: (1/u)·u' = 1·ln x + x·(1/x) = ln x +
1. So u' = u(ln x + 1) = x^x(ln x + 1).
Solution — Step 2: Use product rule for y = u·v
y' = u'·v + u·v'. Substitute u, u', v, v'. We computed u' = x^x(ln x + 1), u = x^x, v = sin x, v' =
cos x.
Therefore: y' = x^x(ln x + 1)·sin x + x^x·cos x Factor x^x: y' = x^x[ (ln x + 1)·sin x + cos x ].
This is the simplified derivative.
Made for Stuvia-style upload — keep previews light and watermarked. Page 1
