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Math · Calculus I · Worked example

Differentiate e^(−x²) with the chain rule

Differentiate y = e^(−x²), then use the derivative to find where the curve rises and falls.

y=e−x2

Name the layers

Work out e^(−x²) for x = 2: first −x² = −(2²) = −4, then e^(−4). The exponent is worked out first, so it is the inside, and raising e to that power is the outside. Note that −x² means −(x²), so it is never positive.

u=−x2,y=eu

Differentiate the outside

eᵘ is its own derivative, so dy/du = eᵘ. The exponent is copied unchanged.

d⁢yd⁢u=eu

Differentiate the inside

The derivative of −x² is −2x.

d⁢ud⁢x=−2⁢x

Multiply, then put the inside back

Multiply the rates, replace u with −x², and write the factor −2x in front.

d⁢yd⁢x=e−x2⋅(−2⁢x)=−2⁢xe−x2

Read the sign of the derivative

e^(−x²) is positive for every x, so the derivative has the same sign as −2x. For x < 0 the derivative is positive and the curve rises. At x = 0 it is 0, the flat top of the bell at y = 1. For x > 0 it is negative and the curve falls.

Result

dy/dx = −2x e^(−x²). The curve rises for x < 0, levels off at its peak (0, 1) and falls for x > 0.

d⁢yd⁢x=−2⁢xe−x2

Your turn

Differentiate y = ln(x² + 1).

Show the answer and explanation

dy/dx = 2x/(x² + 1).

The inside is u = x² + 1 and the outside is ln u, whose derivative is 1/u. Multiply by the inside’s derivative, 2x: (1/(x² + 1))·2x = 2x/(x² + 1). Since x² + 1 is positive for every x, the logarithm and its derivative are defined everywhere.

y=ln(x2+1)d⁢yd⁢x=1x2+1⋅2⁢xd⁢yd⁢x=2⁢xx2+1

Keep exploring

Open the curve and its derivative in Graph: g(x) = −2xe^(−x²) is positive left of 0, crosses zero at the peak x = 0 and is negative to the right.

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