Free Math Worksheets

- Substitute \(1+h\) for \(x\) and expand the square. - A tangent line is determined by its slope and the point of tangency. - Compare how the slope changes on a parabola with how it behaves on a line. - Think about what a derivative function says about slopes at different inputs.

1. Use the limit definition at \(x=1\): \(f'(1)=\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\). 2. Substitute and simplify: \(f'(1)=\lim_{h\to 0}\frac{(1+h)^2-1-(1^2-1)}{h}=\lim_{h\to 0}\frac{2h+h^2}{h}=\lim_{h\to 0}(2+h)=2\). 3. The point of tangency is \((1,f(1))=(1,0)\), and the slope is \(2\). Using point-slope form gives \(y-0=2(x-1)\), so the tangent line is \(y=2x-2\). 4. A linear function has the same slope at every point, so its derivative is constant. The slope of the parabola changes with \(x\), so the derivative values of \(f\) are not constant.

1. \(f'(1)=2\) 2. \(y=2x-2\) 3. A linear function has a constant derivative, while the derivative of \(f(x)=x^2-1\) varies with \(x\).

- Combine the two fractions in the numerator first. - Notice the sign relationship between \(x_0-x\) and \(x-x_0\). - Recall that \(x^{-k}=\frac{1}{x^k}\).

1. Begin with \(\frac{\frac{1}{x}-\frac{1}{x_0}}{x-x_0}\). Combining the numerator gives \(\frac{x_0-x}{xx_0(x-x_0)}\). Since \(x_0-x=-(x-x_0)\), this simplifies to \(-\frac{1}{xx_0}\) for \(x\neq x_0\). Therefore, \(f'(x_0)=\lim_{x\to x_0}-\frac{1}{xx_0}=-\frac{1}{x_0^2}\). 2. The power rule gives \(f'(x)=-x^{-2}=-\frac{1}{x^2}\). 3. Evaluating the power-rule result at \(x_0\) gives \(-\frac{1}{x_0^2}\), exactly matching the limit result.

1) \(f'(x_0)=-\frac{1}{x_0^2}\) 2) \(f'(x)=-x^{-2}=-\frac{1}{x^2}\) 3) Both methods give the same derivative for \(x\neq0\).