Biology Forums - Study Force

Find the derivative of this by the delta method

\(y=x^2+3x-\frac{2}{x}\)

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\left(x^2+3x-\frac{2}{x}\right)\)

Apply \(x-\Delta x\) into where you see x's:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{\left(x+\Delta x\right)^2+3x+3\Delta x-\frac{2}{x+\Delta x}-\left(x^2+3x-\frac{2}{x}\right)}{\Delta x}\)

Distribute the negative where you see: \(-\left(x^2+3x-\frac{2}{x}\right)\)

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{\left(x+\Delta x\right)^2+3x+3\Delta x-\frac{2}{x+\Delta x}-x^2-3x+\frac{2}{x}}{\Delta x}\)

Expand \(\left(x+\Delta x\right)^2\) in the numerator:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{x^2+\left(2x\cdot \Delta x\right)+\Delta x^2+3x+3\Delta x-\frac{2}{x+\Delta x}-x^2-3x+\frac{2}{x}}{\Delta x}\)

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{2x\cdot \Delta x+\Delta x^2+3\Delta x-\frac{2}{x+\Delta x}+\frac{2}{x}}{\Delta x}\)

Combine the fractions found in the numerator:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{2x\cdot \Delta x+\Delta x^2+3\Delta x+\frac{-2x+2\left(x+\Delta x\right)}{\left(x+\Delta x\right)x}}{\Delta x}\)

Now combine all the terms in the numerator:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{\frac{x\left(x+\Delta x\right)\left(2x\cdot \Delta x+\Delta x^2+3\Delta x\right)-2x+2\left(x+\Delta x\right)}{\left(x+\Delta x\right)x}}{\Delta x}\)

Simplify the fractions more:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{x\left(x+\Delta x\right)\left(2x\cdot \Delta x+\Delta x^2+3\Delta x\right)-2x+2\left(x+\Delta x\right)}{x\left(x+\Delta x\right)\Delta x}\)

Expand the numerator:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{\left(x^2+x\cdot \Delta x\right)\left(2x\cdot \Delta x+\Delta x^2+3\Delta x\right)-2x+2x+2\Delta x}{x\left(x+\Delta x\right)\Delta x}\)

Simplify and expand:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{2x^3\Delta x+\Delta x^2x^2+3\Delta x\cdot x^2+2x^2\Delta x^2+\Delta x^3\cdot x+3\Delta x^2\cdot x+2\Delta x}{x\left(x+\Delta x\right)\Delta x}\)

Now give the denominator to each numerator:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{2x^3\Delta x}{x\left(x+\Delta x\right)\Delta x}+\frac{\Delta x^2x^2}{x\left(x+\Delta x\right)\Delta x}+\frac{3\Delta x\cdot x^2}{x\left(x+\Delta x\right)\Delta x}+\frac{2x^2\Delta x^2}{x\left(x+\Delta x\right)\Delta x}+\frac{\left(\Delta x^3\cdot x\right)}{x\left(x+\Delta x\right)\Delta x}+\frac{\left(3\Delta x^2\cdot x\right)}{x\left(x+\Delta x\right)\Delta x}+\frac{2\Delta x}{x\left(x+\Delta x\right)\Delta x}\)

Simplify each fraction:

\(\frac{dy}{dx}=\lim _{\Delta x\rightarrow 0}\ \frac{2x^2}{\left(x+\Delta x\right)}+\frac{\Delta x\cdot x}{\left(x+\Delta x\right)}+\frac{3x}{\left(x+\Delta x\right)}+\frac{4x\cdot \Delta x}{\left(x+\Delta x\right)\Delta x}+\frac{\Delta x^2}{\left(x+\Delta x\right)}+\frac{3\Delta x}{x\left(x+\Delta x\right)}+\frac{2}{x\left(x+\Delta x\right)}\)

Take the limit of EACH fraction:

\(\frac{dy}{dx}=2x+0+3+0+0+0+\frac{2}{x^2}\)

One last simplification:

\(\frac{dy}{dx}=2x+3+\frac{2}{x^2}\)

Finished!