Gradient on curves

Question

Please with a bit of explanation, what is the gradient on the curve $y = 16/x$ where $x = 8$. I'm finding it hard to solve problem like this.

Answer 1

$\bf{My\; Solution::}$ Given $\displaystyle y = \frac{16}{x}\;,$ Then $\displaystyle \frac{dy}{dx} = \frac{d}{dx}\left(\frac{16}{x}\right) = -\frac{16}{x^2}.$

Now $\displaystyle \left(\frac{dy}{dx}\right)_{x=8} = -\left(\frac{16}{x^2}\right)_{x=8} = -\frac{16}{64} = -\frac{1}{4}$.

Now We know That $\displaystyle \bf{Gradient} = \left(\frac{dy}{dx}\right)_{x=8}=-\frac{1}{4}.$

Answer 2

Perhaps some confusion is coming from the fact that there is no well-defined notion of "gradient on a curve". You want the idea of "gradient of a function". If you mean to write $y=f(x)$ then it should be a bit more clear that the gradient is just the derivative $$f'(x)=\frac{dy}{dx}=-\frac{16}{x^2}$$

and $f'(8)=1/4$.

Answer 3

Notice, the derivative $\frac{dy}{dx}$ of any function $y=f(x)$ shows the gradient/slope of the tangent at a general point on the curve

hence we have for given function $y=\frac{16}{x}$

$$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{16}{x}\right)$$$$ =16\frac{d}{dx}\left(x^{-1}\right)$$ $$=16(-x^{-1-1})=16(-x^{-2})=-\frac{16}{x^2}$$ Hence, the slope at $x=8$ is as follows $$\left(\frac{dy}{dx}\right)_{x=8}=\left(\frac{-16}{x^2}\right)_{x=8}$$ $$=\left(\frac{-16}{8^2}\right)=-\frac{16}{64}=-\frac{1}{4}$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{gradient on the curve}=-\frac{1}{4}}}$$