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April 11th, 2012, 02:56 PM   #1
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Derivatives-Help

[attachment=0:2pc1v6ec]Derivatives.docx[/attachment:2pc1v6ec]
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 April 11th, 2012, 03:52 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivatives-Help Here are the problems: A. Find the derivative using the definition of the derivative for the following function: $f(x)=x^2+3$ B. Find the derivative of the given functions: 1. $y=3$ 2. $g(x)=x^2+4$ 3. $h(t)=-2t^2+3t-6$ 4. $y=\sin x$ 5. $h(x)=\frac{1}{3x}$ 6. $f(x)=x^3-3x-2x^{\small{-4}}$ 7. $y=\frac{3x-2}{2x-3}$ 8. $g(x)=$$x^2-2x+1$$$$x^3-1$$$ 9. $f(x)=x^2-\frac{1}{2}\cos x$ 10. $g(x)=\frac{1}{x}+3\sin x$ 11. $y=$$e^{2x^2+1}$$^3$ 12. $f(x)=\sqrt[5]{\frac{x^2+1}{1-x^3}}$ 13. $y=\ln$$x^4+5$$$ C. 1. Find the equation of the tangent line to the function at the given point: $f(x)=(x-1)^2$ at the point $$$-2,9$$$ 2. A 5 meter ladder leaning against a wall has its lower part moving away from the wall at the rate of 1 meter/second. At what speed is the top of the ladder moving down when the lower part of the ladder is 2 meters from the wall? What have you done, and where are you stuck?
April 12th, 2012, 02:45 PM   #3
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Re: Derivatives-Help

I tried to solve some of these derivatives. But I am not sure about the answers.
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 April 12th, 2012, 03:34 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivatives-Help A. Find the derivative using the definition of the derivative for the following function: $f(x)=x^2+3$ You have the correct result, but you show no work, which implies you took the derivative using the power rule. What you are presumably supposed to show is: Begin with the definition of the derivative: $\frac{d}{dx}$$f(x)$$\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ $\frac{d}{dx}$$f(x)$$=\lim_{h\to0}\frac{$$(x+h)^2+3$$-$$x^2+3$$}{h}$ $\frac{d}{dx}$$f(x)$$=\lim_{h\to0}\frac{x^2+2hx+h^2 +3-x^2-3}{h}$ $\frac{d}{dx}$$f(x)$$=\lim_{h\to0}\frac{2hx+h^2}{h}$ $\frac{d}{dx}$$f(x)$$=\lim_{h\to0}2x+h=2x$ B. Find the derivative of the given functions: 1. $y=3$ $y'=0$ Correct. 2. $g(x)=x^2+4$ $g'(x)=2x$ Correct. 3. $h(t)=-2t^2+3t-6$ $h'(t)=-4t+3$ Correct. 4. $y=\sin x$ $y'=\cos x$ Correct. 5. $h(x)=\frac{1}{3x}$ This one you have incorrect. Try writing it as: $h(x)=\frac{1}{3}x^{\small{-3}}$ Now use the power rule: $\frac{d}{dx}$$k\cdot x^r$$=kr\cdot x^{r-1}$ 6. $f(x)=x^3-3x-2x^{\small{-4}}$ Use the power rule term by term. 7. $y=\frac{3x-2}{2x-3}$ You have this one incorrect. Use the quotient rule: $\frac{d}{dx}$$\frac{f(x)}{g(x)}$$=\frac{g(x)f' (x)-g#39;(x)f(x)}{g^2(x)}$ 8. $g(x)=$$x^2-2x+1$$$$x^3-1$$$ You have this one incorrect. Use the product rule: $\frac{d}{dx}$$f(x)\cdot g(x)$$=f(x)\cdot g'(x)+f#39;(x)\cdot g(x)$ 9. $f(x)=x^2-\frac{1}{2}\cos x$ $f'(x)=2x-\frac{1}{2}$$-\sin x$$=2x+\frac{1}{2}\sin x$ Correct. 10. $g(x)=\frac{1}{x}+3\sin x$ Incorrect. You differentiated the trig. term correctly, but think of the first term as $x^{\small{-1}}$ and use the power rule. 11. $y=$$e^{2x^2+1}$$^3$ You may either use the rule for exponents $$$a^b$$^c=a^{bc}$ to rewrite then differentiate, or use the power rule and chain rule where the rule for differentiating powers of e is then used. See if you can attempt this one now. 12. $f(x)=\sqrt[5]{\frac{x^2+1}{1-x^3}}$ Rewrite using a rational exponent: $f(x)=$$\frac{x^2+1}{1-x^3}$$^{\small{\frac{1}{5}}}$ Now use the power rule and chain rule where the quotient rule is then used. 13. $y=\ln$$x^4+5$$$ Use the rule for differentiating the natural log function along with the chain rule where the power rule is then used. C. 1. Find the equation of the tangent line to the function at the given point: $f(x)=(x-1)^2$ at the point $$$-2,9$$$ Use: $y-y_1=f#39;$$x_1$$$$x-x_1$$$ 2. A 5 meter ladder leaning against a wall has its lower part moving away from the wall at the rate of 1 meter/second. At what speed is the top of the ladder moving down when the lower part of the ladder is 2 meters from the wall? Let x represent the distance of the lower part of the ladder from the wall and y represent the position of the top of the ladder on the wall. Use the Pythagorean theorem to relate x and y then implicitly differentiate with respect to time t. Solve for dy/dt and plug in the given values for x, y and dx/dt.
April 13th, 2012, 06:18 PM   #5
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Re: Derivatives-Help

[attachment=0:wr0rhcwh]Derivatives solutions.docx[/attachment:wr0rhcwh]
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 April 13th, 2012, 06:56 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivatives-Help You did very well! Just a few comments: Section B 10.) Correct, but bring cos(x) down from the exponent. Going by your working, I'm sure this is just a typo. 11.) Correct, but you might want to clean up your working some, it is a little hard to follow. 12.) Correct, but you could simplify the end result a little. That is up to you. Section C 2.) Your result is not correct. You want to use the Pythagorean theorem to state: $x^2+y^2=5^2$ Implicitly differentiate with respect to time t: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ Solve for $\frac{dy}{dt}$: $\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}$ We are told: $\frac{dx}{dt}=1\text{ \frac{m}{s}}$ $x=2\text{ m}$ $y=\sqrt{5^2-2^2}=\sqrt{21}\text{ m}$ and so we have: $\frac{dy}{dt}=-\frac{2\text{ m}}{\sqrt{21}\text{ m}}$$1\text{ \frac{m}{s}}$$=-\frac{2}{\sqrt{21}}\:\text{\frac{m}{s}}$ The negative sign means the top of the ladder is moving down, and since the questions asks for the downward speed, we may say the top of the ladder is moving down at a a speed of $\frac{2}{\sqrt{21}}\:\text{\frac{m}{s}}$.

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