Diagnose and fix common derivative calculation errors with step-by-step solutions
Function: f(x) = sin(x²)
Student's answer: f'(x) = cos(x²)
f(x) = sin(x²)
f'(x) = cos(x²) · 2x
f'(x) = 2x cos(x²)
The student correctly identified the outer function (sin) but forgot to multiply by the derivative of the inner function (x²).
Function: f(x) = (sin x)²
Student's answer: f'(x) = 2x sin(x)
f(x) = (sin x)² = sin²(x)
Let u = sin x, then f(u) = u²
f'(x) = 2u · u' = 2 sin x · cos x
f'(x) = 2 sin x cos x = sin(2x)
Function: f(x) = x · sin(x)
Student's answer: f'(x) = 1 + cos(x)
Let u = x, v = sin(x)
u' = 1, v' = cos(x)
f'(x) = u'v + uv'
f'(x) = 1·sin(x) + x·cos(x)
f'(x) = sin(x) + x cos(x)
The student added the derivatives instead of using the product rule formula.
"Derivative of First times Second,
Plus First times Derivative of Second"
Function: f(x) = (x² + 1)/(x - 1)
Student's answer: f'(x) = [2x·(x-1) + (x²+1)·1] / (x-1)²
Let u = x² + 1, v = x - 1
u' = 2x, v' = 1
f'(x) = (u'v - uv') / v²
f'(x) = [2x·(x-1) - (x²+1)·1] / (x-1)²
f'(x) = (2x² - 2x - x² - 1) / (x-1)²
f'(x) = (x² - 2x - 1) / (x-1)²
"Low D-High minus High D-Low,
Over the Square of What's Below"
Function: f(x) = 3x² - 4x + 5
Student's answer: f'(x) = 6x - 4 + 0 = 6x - 4
But then: f''(x) = 6 - 0 = 6
Error: Forgot negative sign in second derivative
f(x) = 3x² - 4x + 5
f'(x) = 6x - 4
f''(x) = 6
Note: -4 differentiates to 0, not -0
When differentiating polynomials, handle each term separately. Constants become 0, and coefficients carry through with proper signs.
Function: f(x) = e^(2x + 3)
Student's answer: f'(x) = e^2x + 3 · 2
f(x) = e^(2x + 3)
Let u = 2x + 3
f'(x) = e^(2x + 3) · 2
f'(x) = 2e^(2x + 3)
Always use parentheses when the exponent contains more than one term. e^(2x+3) is different from e^2x + 3.
Function: f(x) = (x² + 1)³
Given solution: f'(x) = 3(x² + 1)² · 2x = 6x(x² + 1)²
Is this correct?
Function: f(x) = x · e^x
Given solution: f'(x) = 1 · e^x = e^x
What's wrong?
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