Free math worksheets

- A derivative at a point uses a difference quotient whose input change approaches zero. - The numerator should compare the function value at the base point with a nearby function value. - Check that the denominator is exactly the same input change used in the nearby input.

1. The derivative at \(x=3\) is defined by comparing \(f(3+h)\) with \(f(3)\) and dividing by the input change \(h\). 2. The input change must approach \(0\). 3. Therefore, expression A is the derivative definition for \(f'(3)\).

A) \(\lim_{h\to0}\frac{f(3+h)-f(3)}{h}\)

- Keep the base input fixed at the point where the derivative is requested. - The nearby input differs from the base input by \(h\). - The denominator is the same change \(h\), which approaches \(0\).

1. Use the base input \(a=-2\) in the derivative definition \(f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\). 2. Substituting \(a=-2\) gives \(f'(-2)=\lim_{h\to0}\frac{f(-2+h)-f(-2)}{h}\).

\(f'(-2)=\lim_{h\to0}\frac{f(-2+h)-f(-2)}{h}\)

- Substitute the function into the difference quotient before simplifying. - Expand the squared binomial and look for a common factor of \(h\). - Cancel only for nonzero \(h\), then take the limit.

1. Start with \(f'(1)=\lim_{h\to0}\frac{(1+h)^2-1}{h}\). 2. Expand the numerator: \((1+h)^2-1=2h+h^2\). 3. For \(h\ne0\), the quotient simplifies to \(2+h\). 4. Taking the limit gives \(f'(1)=2\).

\(f'(1)=2\)

- Substitute the base input into the alternate derivative definition. - Direct substitution gives an indeterminate form. What algebraic identity can turn a difference of square roots into a factor of \(x-4\)? - After the factor causing the indeterminate form is removed, evaluate the limit.

1. Substitute \(a=4\): \(f'(4)=\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}\). 2. Multiply by the conjugate: \(\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{x-4}{(x-4)(\sqrt{x}+2)}\). 3. For \(x\ne4\), the quotient simplifies to \(\frac{1}{\sqrt{x}+2}\). 4. Therefore, \(f'(4)=\frac{1}{4}\).

\(f'(4)=\frac{1}{4}\)

- 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 \(x\).

- Evaluate the difference quotient separately for positive and negative values of \(h\). - The absolute-value expression simplifies differently on the two sides of zero. - A two-sided derivative exists only when the one-sided derivative limits agree.

1. For \(h>0\), \(\frac{f(h)-f(0)}{h}=\frac{|h|}{h}=1\), so the right-hand derivative is \(1\). 2. For \(h<0\), \(\frac{f(h)-f(0)}{h}=\frac{|h|}{h}=-1\), so the left-hand derivative is \(-1\). 3. The one-sided derivative limits are different, so the two-sided derivative at \(0\) does not exist.

\(f\) is not differentiable at \(x=0\); the left-hand derivative is \(-1\) and the right-hand derivative is \(1\).

- Identify the input that must remain fixed in a derivative at a point. - Compare the student's numerator with the general \(h\to0\) derivative definition. - After correcting the setup, expand the cube and simplify before taking the limit.

1. The second function value must stay at the fixed base input, so the numerator should contain \(f(1)\), not \(f(h)\). 2. The correct limit is \(\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{(1+h)^3-1}{h}\). 3. Expanding gives \(\frac{3h+3h^2+h^3}{h}=3+3h+h^2\) for \(h\ne0\). 4. Taking the limit gives \(f'(1)=3\).

The error is using \(f(h)\) instead of the fixed value \(f(1)\). The correct limit is \(\lim_{h\to0}\frac{f(1+h)-f(1)}{h}\), and its value is \(3\).

- Substitute the reciprocal function directly into the difference quotient. - Combine the two fractions in the numerator before dividing by \(h\). - Look for a factor of \(h\) that cancels before taking the limit.

1. Start with \(f'(a)=\lim_{h\to0}\frac{\frac{1}{a+h}-\frac{1}{a}}{h}\). 2. Combine the fractions in the numerator: \(\frac{1}{a+h}-\frac{1}{a}=\frac{-h}{a(a+h)}\). 3. For \(h\ne0\), the difference quotient simplifies to \(-\frac{1}{a(a+h)}\). 4. Taking the limit gives \(f'(a)=-\frac{1}{a^2}\).

\(f'(a)=-\frac{1}{a^2}\), for \(a\ne0\)

- Read the approximate y-coordinates of the two marked points from the graph. - Use rise over run with \(P\) as the fixed point for each secant. - The derivative is the common value approached by secant slopes as the moving point approaches \(P\).

1. From the graph, \(\sin(0.5)\approx0.48\) and \(\sin(-0.5)\approx-0.48\). 2. The left secant slope is approximately \(\frac{0-(-0.48)}{0-(-0.5)}\approx0.96\). 3. The right secant slope is approximately \(\frac{0.48-0}{0.5-0}\approx0.96\). 4. Both nearby secants suggest a limiting tangent slope near \(1\) at \(P\). 5. As the second point approaches \(P\) from either side, the secant slopes should approach the same limiting value.

a) Each secant slope is approximately \(0.96\). b) The limiting tangent slope is approximately \(1\). c) The secant slopes should approach the same value as the second point approaches \(P\).

- 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 \(x\neq0\).