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Differentials

Differentials represent infinitesimally small changes in a variable. This concept is directly related to our study of derivatives.

Increments

For a function f(x)f(x), the increment, denoted Δy\Delta y, is the true change in yy when the independent variable varies by Δx=x1−x0\Delta x = x_1 - x_0.

Δy=f(x1)−f(x0)=f(x0+Δx)−f(x0)\boxed{\Delta y = f(x_1) - f(x_0) = f(x_0 +\Delta x) - f(x_0)}

Differentials

For a function f(x)f(x) , dxdx and dydy are called differentials and are related by

dy=f′(x)dx\boxed{dy=f'(x)dx}

Approximations with Differentials

When Δx\Delta x , the change in xx, is small, increments can be approximated by differentials.

Δy≈dy=f′(x)dx\boxed{\Delta y \approx dy = f'(x)dx}


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Example: Differentials

Find an approximation for the percentage change in the area of a circle if its radius is decreased by 2%.

Remember dA≈ΔAdA\approx \Delta A and dr≈Δrdr \approx \Delta r

Percentage Change in r:Δrr=−.02\displaystyle \text{Percentage Change in }r: \\\frac{\Delta r}{r}=-.02


Area of Circle:A=πr2\text{Area of Circle}:\\ A=\pi r^2

  ⟹  dA=2πr∗dr\implies dA=2\pi r*dr

Approximation of change in A:ΔA≈dA=2πrdr≈2πrΔr\displaystyle\text{Approximation of change in }A: \\\Delta A \approx dA=2\pi rdr\approx2\pi r\Delta r


Percentage Change in A:ΔAA≈2πrΔrπr2=2∗Δrr=2(−.02)=−.044% decrease\displaystyle \text{Percentage Change in }A: \\\frac{\Delta A}{A}\approx \frac{2\pi r\Delta r}{\pi r^2}\\ \text{} \\=2*\frac{\Delta r}{r} \\ \text{} \\=2(-.02) \\ \text{} \\=-.04\\ \text{}\\ \boxed{4\% \text{ decrease}}










y=Cxny = Cx^n for constants CC and n≠−1n \neq -1. By approximately what percentage will yy change if xx increases by r%r\%?

Extra Practice