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Thread: Find the slope of the tangent line on f at x=1

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    Find the slope of the tangent line on f at x=1

    So I'm studying for my first calculus exam, and this problem has me stumped for hours now.
    Find the slope of the tangent line on f at x=1-f0a351edf18279e0d30a0aada53bbb64.png
    I've used the formula of lim h-->0 f(x+h)-f(x) / h

    It becomes so complicated and confusing as I try to multiply by the conjugate to remove the radicals in the numerator, it just never simplifies to a normal answer. If anybody could help me I'd greatly appreciate it. Thanks!
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    Re: Find the slope of the tangent line on f at x=1

    Quote Originally Posted by anneheather98 View Post
    So I'm studying for my first calculus exam, and this problem has me stumped for hours now.
    Click image for larger version. 

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    I've used the formula of lim h-->0 f(x+h)-f(x) / h

    It becomes so complicated and confusing as I try to multiply by the conjugate to remove the radicals in the numerator, it just never simplifies to a normal answer. If anybody could help me I'd greatly appreciate it. Thanks!
    You don't need to use the definition. Have you learnt the product rule?
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    Re: Find the slope of the tangent line on f at x=1

    How would you use that for the slope? you would find the limit x->1 of 4√x * lim x->1 e^x ? it would just be e then?
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    Re: Find the slope of the tangent line on f at x=1

    The product rule states that: If y =uv, then y'=uv'+ vu'.

    Have you learnt that yet?
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    Re: Find the slope of the tangent line on f at x=1

    Quote Originally Posted by Debsta View Post
    The product rule states that: If y =uv, then y'=uv'+ vu'.

    Have you learnt that yet?
    Never seen that, can you explain the process of getting the slope? I can infer the rest by following the steps you take. While doing these problems we were instructed to use the definition, I don't know if this one is any different as my instructor didn't give us the solution to this one.
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    Re: Find the slope of the tangent line on f at x=1

    Quote Originally Posted by anneheather98 View Post
    Never seen that, can you explain the process of getting the slope? I can infer the rest by following the steps you take. While doing these problems we were instructed to use the definition, I don't know if this one is any different as my instructor didn't give us the solution to this one.
    The "formula" for the slope of a function (ie the slope of the tangent line) is given by the derivative. So you need to find the derivative and then substitute x=1.
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    Re: Find the slope of the tangent line on f at x=1

    Can you please post a solution to a similar question that your instructor has provided you with? This way I can see how you are expected to do this problem. There are several methods.
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    Re: Find the slope of the tangent line on f at x=1

    Quote Originally Posted by anneheather98 View Post
    So I'm studying for my first calculus exam, and this problem has me stumped for hours now.
    Click image for larger version. 

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Views:	21 
Size:	14.6 KB 
ID:	38977
    Comment: If this is the first calculus test and you have not studied the product rule for derivatives then the above question is totally inappropriate.
    If you have not done the product rule, you have a complaint.


    Quote Originally Posted by anneheather98 View Post
    It becomes so complicated and confusing as I try to multiply by the conjugate to remove the radicals in the numerator, it just never simplifies to a normal answer. If anybody could help me I'd greatly appreciate it.
    The slope of $f(x)=\sqrt[4]xe^x$ at $x=1$ is $f'(1)$.

    Using the product rule we get $f'(x)=\left( {\frac{1}{4}{x^{ - \frac{3}{4}}}} \right){e^x} + \sqrt[4]{x}\left( {{e^x}} \right)$
    So $f'(1)=~?$
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    Re: Find the slope of the tangent line on f at x=1

    to use the definition of the slope at $\displaystyle x=1$ write

    $\displaystyle \frac{f(1+h)-f(1)}{ h}=e\frac{e^h-1}{h}+e^{h+1}\frac{(1+h)^{1/4}-1}{h}$

    now take the limit as $\displaystyle h \to 0$
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    Re: Find the slope of the tangent line on f at x=1

    Quote Originally Posted by Idea View Post
    to use the definition of the slope at $\displaystyle x=1$ write

    $\displaystyle ... = \ e\frac{e^h-1}{h}+e^{h+1}\frac{(1+h)^{1/4}-1}{h}$

    now take the limit as $\displaystyle h \to 0$
    Idea, with your method, how would you expect beginning calculus student anneheather98 to deal
    with the two terms above that are indeterminate forms as you state to "now take the limit as h --> 0?"

    You agree that at this point it would preclude using the product rule or L'Hopital's Rule, because it is
    by the definition?
    Last edited by greg1313; Sep 29th 2018 at 10:21 AM.
    Thanks from Plato
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    Re: Find the slope of the tangent line on f at x=1

    you are right!

    I was thinking something like this

    $\displaystyle \frac{e^h-1}{h}\to 1$

    since by definition this is the derivative of $\displaystyle e^x$ at $\displaystyle x=0$

    and for the second limit rationalize the numerator

    $\displaystyle \frac{(1+h)^{1/4}-1}{h}=\frac{1}{\left((1+h)^{1/4}+1\right) \left((1+h)^{1/2}+1\right)}\to \frac{1}{4}$
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