Free math worksheets

- Differentiate the constant multiple without needing a formula for \(f\). - A constant term contributes zero to the derivative. - Use the supplied derivative value only after applying the rule.

1. By the constant-multiple and constant rules, \(h'(x)=4f'(x)\). 2. Therefore, \(h'(3)=4\cdot2=8\).

\(h'(3)=8\)

- The derivative of a sum is the sum of the derivatives. - The original function values are not needed for this calculation. - Keep the sign of the second derivative value when combining them.

1. By the sum rule, \(h'(x)=u'(x)+v'(x)\). 2. Thus, \(h'(2)=3+(-5)=-2\).

\(h'(2)=-2\)

- What is the derivative rule for a sum? - Differentiate the two functions separately. - Compare the sum of the derivatives with their product. - Evaluate both expressions at a simple input to show they are different.

1. The correct sum rule is \(f'(x)=g'(x)+h'(x)\). 2. For the example, \(g'(x)=3x^2\) and \(h'(x)=2x\). 3. Therefore, the correct derivative is \(f'(x)=3x^2+2x\). 4. The student's rule would give \(g'(x)h'(x)=(3x^2)(2x)=6x^3\). 5. These expressions are not equal in general. For example, at \(x=1\), they give \(5\) and \(6\), respectively.

The conjecture is false. The sum rule gives \(f'(x)=g'(x)+h'(x)\). For the example, the correct derivative is \(3x^2+2x\), while the product of the derivatives is \(6x^3\).

- Distinguish multiplication from addition. - What does the constant-multiple rule say? - What happens to a constant coefficient multiplying a power of \(x\)? - Could the graph of \(5x^4\) have slope \(0\) everywhere?

1. The student is confusing a constant term with a constant factor. The number \(5\) multiplies the variable expression; it is not a separate constant term. 2. By the constant-multiple rule, a constant factor remains: \(\frac{d}{dx}[c\,g(x)]=c\,g'(x)\). 3. Since \(\frac{d}{dx}(x^4)=4x^3\), the derivative is \(f'(x)=5(4x^3)=20x^3\).

The student incorrectly differentiated a constant factor as though it were a separate constant term. The correct derivative is \(f'(x)=20x^3\).

- Recall the meaning of a zero exponent. - What happens when an expression is multiplied by \(0\)? - Is \(x^{-1}\) defined at every real value of \(x\)?

1. For \(x\neq0\), \(1=x^0\). 2. Applying the power rule with \(n=0\) gives \(\frac{d}{dx}(x^0)=0x^{-1}=0\) for \(x\neq0\). 3. This agrees with the constant rule. The power-rule expression \(x^{-1}\) is undefined at \(x=0\), so this particular argument does not apply there. However, the original constant function is differentiable at \(0\), and its derivative there is also \(0\).

1) \(f(x)=x^0\) for \(x\neq0\) 2) \(f'(x)=0x^{-1}=0\) for \(x\neq0\) 3) The power-rule argument excludes \(x=0\), but the constant function itself still satisfies \(f'(0)=0\).

- Rewrite reciprocals and square roots as powers. - Differentiate each term separately. - Treat \(k\) as a constant. - Check where the original expressions and their derivatives are defined.

1. Differentiate each term in part a: \(\frac{d}{dx}(6x^{-1})=-6x^{-2}\), \(\frac{d}{dx}(4x^5)=20x^4\), and \(\frac{d}{dx}(-3\sin x)=-3\cos x\). Thus, \(f'(x)=-\frac{6}{x^2}+20x^4-3\cos x\), for \(x\ne0\). 2. For part b, \(\frac{d}{dx}(10x^{1/2})=5x^{-1/2}=\frac{5}{\sqrt{x}}\), \(\frac{d}{dx}\left(-\frac{1}{2}\cos x\right)=\frac{1}{2}\sin x\), and the derivative of \(12\) is \(0\). Therefore, \(f'(x)=\frac{5}{\sqrt{x}}+\frac{1}{2}\sin x\), for \(x>0\). 3. In part c, treat \(k\) as a constant: \(f'(x)=k\cos x-\frac{2}{x^2}\), for \(x\ne0\).

a) \(f'(x)=-\frac{6}{x^2}+20x^4-3\cos x\), for \(x\ne0\) b) \(f'(x)=\frac{5}{\sqrt{x}}+\frac{1}{2}\sin x\), for \(x>0\) c) \(f'(x)=k\cos x-\frac{2}{x^2}\), for \(x\ne0\)

- What is the derivative of a constant? - Substitute the two factors into the product rule. - What happens to the first product-rule term when one factor is constant? - Which method requires fewer steps?

1. By the constant multiple rule, \(f'(x)=8\cos(x)\). 2. Using the product rule, \(u'(x)=0\) and \(v'(x)=\cos(x)\), so \(f'(x)=u'(x)v(x)+u(x)v'(x)\) \(=0\sin(x)+8\cos(x)=8\cos(x)\). 3. The methods agree because the derivative of the constant factor is zero. The constant multiple rule is more efficient because it omits the product-rule term that equals zero.

1. \(f'(x)=8\cos(x)\) 2. \(f'(x)=8\cos(x)\). The product rule gives the same result because the derivative of \(8\) is \(0\); the constant multiple rule is more efficient because it avoids writing the zero product-rule term.

- Differentiate each term while treating \(a\) as a constant coefficient. - Use the derivative value at the specified input to create an equation for \(a\). - Solve the resulting linear equation only after differentiation.

1. Differentiate using the constant-multiple and sum rules: \(p'(x)=3ax^2+2\). 2. Substitute \(x=1\): \(3a+2=11\). 3. Solving gives \(a=3\).

\(a=3\)

- Read both y-values at \(x=4\). - Use rise over run on each dashed tangent segment. - Compare \(\frac{g'(4)}{f'(4)}\) with \(\frac{g(4)}{f(4)}\).

1. From the graph, \(f(4)=2\) and \(g(4)=4\). Since \(g(4)=kf(4)\), \(4=2k\), so \(k=2\). 2. The tangent to \(f\) rises \(2\) units over a run of \(2\), so \(f'(4)=1\). The tangent to \(g\) rises \(4\) units over a run of \(2\), so \(g'(4)=2\). 3. Therefore, \(g'(4)=2f'(4)=kf'(4)\). This illustrates the constant multiple rule: if \(g(x)=kf(x)\), then \(g'(x)=kf'(x)\).

1. \(k=2\) 2. \(f'(4)=1\) and \(g'(4)=2\) 3. \(g'(4)=kf'(4)\), illustrating the constant multiple rule.

- Read the derivative of \(u\) from its tangent line at the marked input. - Because \(v\) is a line, its derivative is its constant slope. - Apply the difference rule after obtaining the two component slopes.

1. The tangent line \(t\) to \(u\) passes through grid points \((0.5,0)\) and \((1,1)\), so \(u'(1)=\frac{1-0}{1-0.5}=2\). 2. The graph of \(v\) is a line with slope \(-1\), so \(v'(1)=-1\). 3. By the difference rule, \(w'(1)=u'(1)-v'(1)=2-(-1)=3\).

\(w'(1)=3\)