0:00 / 0:00

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=x1x0\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.

Δydy=f(x)dx\boxed{\Delta y \approx dy = f'(x)dx}


0:00 / 0:00

Example: Differentials

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

Remember dAΔAdA\approx \Delta A and drΔrdr \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=2πrdr\implies dA=2\pi r*dr

Approximation of change in A:ΔAdA=2πrdr2πrΔr\displaystyle\text{Approximation of change in }A: \\\Delta A \approx dA=2\pi rdr\approx2\pi r\Delta r


Percentage Change in A:ΔAA2πrΔrπr2=2Δrr=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 n1n \neq -1. By approximately what percentage will yy change if xx increases by r%r\%?

Extra Practice