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

  • 0
    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.2012-04-28

2 Answers 2

1

I reckon you are having trouble finding

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

It's

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

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 $k=\frac{-8}{3}$

  • 0
    @SbSangpi , hope that helps.2012-04-29
1

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.

Steps:

  • 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 $k$.

  • 0
    @Sb Sangpi: Your derivative is wrong. Try writing your function as $y = k(2x^3 + 2)^{-1}$ and use the chain rule.2012-04-28