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# Can someone find the slope of the tangent lin

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2/10/2015 4:39:50 PM Posted: 2 years ago Not enough room in the topic title to put what I was asking for but here it is...
f(x)=1/(square root of(3x+1)) Can someone find the slope and provide work? Thank you. |

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2/10/2015 6:02:13 PM Posted: 2 years ago At 2/10/2015 4:39:50 PM, MasturDbtor wrote: Why does everyone have to be in differential calculus...it makes me feel dumb. ;_; |

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2/11/2015 1:06:52 AM Posted: 2 years ago At 2/10/2015 4:39:50 PM, MasturDbtor wrote: Lol, I took this back in high school, but strangely enough, I'm taking differential equations in college right now (after Calc 3, strangely) and just did one of these the other day. This'll probably be a bit messy, but I'll try it and write it out. You may want to try, for future reference, wolframalpha.com. They provide step-by-step solutions (or just solutions - which you may want, since paying is a bit silly) to differential and integration problems. I'm obviously not suggesting to use it in place of doing out the problems, but it's a good way to check your answers. ~ResponsiblyIrresponsibleDDO's Economics Messiah |

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2/11/2015 1:09:32 AM Posted: 2 years ago On second thought.....I should be doing my own math homework and this was posted 8 hours ago, lol.
That, and I started taking the derivative and got scared away by the massive chain rule, lol. ~ResponsiblyIrresponsibleDDO's Economics Messiah |

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2/11/2015 9:12:17 AM Posted: 2 years ago At 2/10/2015 4:39:50 PM, MasturDbtor wrote: Do your own homework "I will.....not suck your dick," Sui_generis |

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2/11/2015 6:04:58 PM Posted: 2 years ago I won't give the answer, because that defeats the purpose of homework, but I will elaborate on the process. You just take the derivative of the function (as you do for pretty much everything in basic Calc). A tangent line is just one that runs through a given point with the slope the function has at that point, so you just plug the x-coordinate into the derivative and BAM!
SO to Bailey, the love of my life <3 Believer in the heart of the cards. " . . . most people can't see the cherries for the trees" - Mixed metaphors are bae |

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2/11/2015 7:49:33 PM Posted: 2 years ago Note that W30;(3x + 1) = (3x + 1)^(")
By chain rule dy/dx = dy/du * du/dx: u = 3x + 1 du/dx = 3 y = u^(") dy/du = 1/(2u^(")) f'(x) = 3/(2u^(")) => 3/(2W30;(3x + 1)) Hence, f'(a) = 3/(2W30;(3a + 1)) |