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The graph of the function with rule $\frac{k}{2(x^3+1)}$ has gradient 1 when $x=1$.Find the value of k
the answer is $\frac{-8}{3}$

I did it
find derivative of
sub $x=1$
$\frac{k}{2(x^3+1)}$ $x=6$
$k=12$ still can't get $\frac{-8}{3}$
help me out thanks.

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Your derivative with respect to $x$ is wrong. – J. M. Apr 28 '12 at 13:09
edited mistake, still wrong? – Sb Sangpi Apr 28 '12 at 13:13
You still haven't given the derivative with respect to $x$, you still seem to differentiate with respect to $k$, and that's where your trouble is starting... – J. M. Apr 28 '12 at 13:16
edited, still wrong? – Sb Sangpi Apr 28 '12 at 13:54
Well, if you don't get the answer -$\frac{8}{3}$, then it's going to be wrong, yes. I suggest you revise how to find the derivative of functions, e.g., how would you differentiate $y = \frac{1}{x}$ with respect to $x$? once you feel comfortable doing this, attempt the question again. – stariz77 Apr 28 '12 at 14:08
up vote 1 down vote accepted

I reckon you are having trouble finding



$$\frac{k}{2} \cdot \left(\frac{d}{dx}(x^3+1)^{-1}\right) = \frac{k}{2} \cdot (-1) \cdot (x^3+1)^{-2} \cdot (3x^2) = \frac{-3kx^2}{2(x^3+1)^2}.$$

That's the correct way to find derivatives, or you could simply use the quotient rule.

Now , its given that the value of derivative is 1 when x=1, so we have,

$\frac{-3k}{8}=1$ this gives $k=\frac{-8}{3}$

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still $x=1$ is $\frac{-3k}{4}$ i can't get $-\frac{8}{3}$ – Sb Sangpi Apr 29 '12 at 9:25
@SbSangpi , hope that helps. – Tomarinator Apr 29 '12 at 12:09

You need to take the derivative of $\frac{k}{2(x^3 + 1)}$ with respect to $x$ first before you substitute in $x = 1$ and set the equation equal to 1.


  • Find the derivative of your function with respect to $x$ in order to find its gradient at $x$.

  • Use the given values for $x$ and the gradient.

  • Solve for $k$.

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edited, still can't get answer! thx – Sb Sangpi Apr 28 '12 at 13:48
@Sb Sangpi: Your derivative is wrong. Try writing your function as $y = k(2x^3 + 2)^{-1}$ and use the chain rule. – stariz77 Apr 28 '12 at 13:59

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