Question

use the derivative shortcuts from this chapter to calculate the derivatives. either show all your work, or explain what shortcut you are using.

1. figure out:

a) $\frac{d}{dx}(4 + 5)$ b) $(5x^{7})$ c) $\frac{d}{dx}(6 + 7x+8x^{2})$ d) $\frac{d}{dx}(6 + 7t+8t^{2})$ e) $(-\frac{1}{x^{2}})$

1. find the derivative of each of the following functions two different (but correct) ways. first use the product and/or quotient rule as it is set up. then simplify first and use only the constant multiple, sum, difference or power rules as needed. you should get the same answer both ways!

a) $x(3x - 17)$ b) $xsqrt{x}$ c) $\frac{x^{2}}{sqrt3{x}}$

1. consider the graph of a function $f$. recall that the slope of the tangent line at a point $x = a$ is equal to the instantaneous rate of change at $x = a$. the equation of the tangent line to the graph of a function $f$ at $x = 4$ is $y=3x - 17$.

(a) what is $f(4)$? what is $f(4)$?
(b) given that $f(x)=ax^{3}+b$, find the constants $a$ and $b$.

Explanation:

Step1: Solve 1.a

The derivative of a constant is 0. Since $4 + 5=9$ (a constant), $\frac{d}{dx}(4 + 5)=0$.

Step2: Solve 1.b

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $y = 5x^{7}$, $y'=5\times7x^{7-1}=35x^{6}$.

Step3: Solve 1.c

Use the sum - rule $\frac{d}{dx}(u + v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)$ and power - rule. $\frac{d}{dx}(6 + 7x+8x^{2})=\frac{d}{dx}(6)+\frac{d}{dx}(7x)+\frac{d}{dx}(8x^{2})=0 + 7+16x=7 + 16x$.

Step4: Solve 1.d

Here the variable of differentiation is $x$ and $t$ is treated as a constant. $\frac{d}{dx}(6 + 7t+8t^{2}) = 0$ since $6 + 7t+8t^{2}$ is a constant with respect to $x$.

Step5: Solve 1.e

Rewrite $y=-\frac{1}{x^{2}}=-x^{-2}$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y'=-(-2)x^{-2 - 1}=2x^{-3}=\frac{2}{x^{3}}$.

Step6: Solve 2.a (Product - rule first)

Let $u = x$ and $v=3x - 17$. By the product - rule $\frac{d}{dx}(uv)=u'v+uv'$, where $u' = 1$ and $v'=3$. So $\frac{d}{dx}(x(3x - 17))=1\times(3x - 17)+x\times3=3x-17 + 3x=6x-17$.
Simplify first: $x(3x - 17)=3x^{2}-17x$. Then $\frac{d}{dx}(3x^{2}-17x)=\frac{d}{dx}(3x^{2})-\frac{d}{dx}(17x)=6x-17$.

Step7: Solve 2.b (Product - rule first)

Rewrite $x\sqrt{x}=x\times x^{\frac{1}{2}}=x^{\frac{3}{2}}$. Let $u = x$ and $v=x^{\frac{1}{2}}$. By the product - rule $\frac{d}{dx}(uv)=u'v+uv'$, $u' = 1$ and $v'=\frac{1}{2}x^{-\frac{1}{2}}$. So $\frac{d}{dx}(x\sqrt{x})=1\times x^{\frac{1}{2}}+x\times\frac{1}{2}x^{-\frac{1}{2}}=x^{\frac{1}{2}}+\frac{1}{2}x^{\frac{1}{2}}=\frac{3}{2}x^{\frac{1}{2}}$. Using the power - rule directly on $y = x^{\frac{3}{2}}$, $y'=\frac{3}{2}x^{\frac{3}{2}-1}=\frac{3}{2}x^{\frac{1}{2}}$.

Step8: Solve 2.c (Quotient - rule first)

Rewrite $\frac{x^{2}}{\sqrt[3]{x}}=x^{2}\times x^{-\frac{1}{3}}=x^{\frac{5}{3}}$. Let $u = x^{2}$ and $v=x^{\frac{1}{3}}$. By the quotient - rule $\frac{d}{dx}(\frac{u}{v})=\frac{u'v - uv'}{v^{2}}$, $u' = 2x$ and $v'=\frac{1}{3}x^{-\frac{2}{3}}$. $\frac{d}{dx}(\frac{x^{2}}{\sqrt[3]{x}})=\frac{2x\times x^{\frac{1}{3}}-x^{2}\times\frac{1}{3}x^{-\frac{2}{3}}}{x^{\frac{2}{3}}}=\frac{2x^{\frac{4}{3}}-\frac{1}{3}x^{\frac{4}{3}}}{x^{\frac{2}{3}}}=\frac{\frac{5}{3}x^{\frac{4}{3}}}{x^{\frac{2}{3}}}=\frac{5}{3}x^{\frac{2}{3}}$. Using the power - rule directly on $y = x^{\frac{5}{3}}$, $y'=\frac{5}{3}x^{\frac{5}{3}-1}=\frac{5}{3}x^{\frac{2}{3}}$.

Step9: Solve 3.a

If the equation of the tangent line to the graph of $y = f(x)$ at $x = 4$ is $y=3x - 17$, then $f(4)$ is the $y$ - value of the tangent line at $x = 4$. Substitute $x = 4$ into $y=3x - 17$, we get $f(4)=3\times4-17=-5$. The slope of the tangent line is $f'(4)$, and the slope of the line $y = 3x - 17$ is 3, so $f'(4)=3$.

Step10: Solve 3.b

Since $f(x)=ax^{3}+b$, then $f'(x)=3ax^{2}$. We know that $f'(4)=3$, so $3a\times4^{2}=3$, $48a = 3$, $a=\frac{1}{16}$. Also, $f(4)=-5$, so $a\times4^{3}+b=-5$. Substitute $a=\frac{1}{16}$ into it: $\frac{1}{16}\times64 + b=-5$, $4 + b=-5$, $b=-9$.

Answer:

1.a) $0$
1.b) $35x^{6}$
1.c) $7 + 16x$
1.d) $0$
1.e) $\frac{2}{x^{3}}$
2.a) $6x-17$
2.b) $\frac{3}{2}x^{\frac{1}{2}}$
2.c) $\frac{5}{3}x^{\frac{2}{3}}$
3.a) $f(4)=-5$, $f'(4)=3$
3.b) $a=\frac{1}{16}$, $b=-9$