Find K such that the line is tangent to $y=-\frac{3}{4}x+4$ The function is $f(x)= \frac{k}{x}$ and the line is $y =-\frac{3}{4}x+3$
So the derivative of $f(x)$ should equal $-\frac{3}{4}$
If I take $k$ to be a constant I can remove it from the fraction like so 
$$f'(x)=k \frac{d}{dx} \frac{1}{x}$$
to get $-\frac{k}{x^2}$ but then I try to solve for $k$
$$-\frac{k}{x^2}=-\frac{3}{4}$$ 
and this is where I get stuck. So should I not even remove $k$ from the differentiation or have I gone in a completely wrong direction?

From the given equation we obtain
$$-\frac{3}{4}x+3=\frac{k}{x}$$ so we get
$$-\frac{3}{4}x^2+3x-k=0$$ solving for $x$ we get
$$x=2\pm \frac{2}{3}\sqrt{9-3k}$$ since the line is a tangent line, we get
$$9-3k=0$$
