Question

consider the following.
$y = \frac{t}{x^{8}}+\frac{x}{t}$
find $\frac{dy}{dx}$.
$\frac{dy}{dx}=\square$
find $\frac{dy}{dt}$.
$\frac{dy}{dt}=\square$

Explanation:

Step1: Rewrite the function for $\frac{dy}{dx}$

We have $y=\frac{t}{x^{8}}+\frac{x}{t}=tx^{-8}+\frac{1}{t}x$. Treat $t$ as a constant.

Step2: Differentiate with respect to $x$

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $\frac{dy}{dx}=- 8tx^{-9}+\frac{1}{t}=-\frac{8t}{x^{9}}+\frac{1}{t}$.

Step3: Rewrite the function for $\frac{dy}{dt}$

$y = \frac{t}{x^{8}}+\frac{x}{t}=\frac{1}{x^{8}}t+xt^{-1}$. Treat $x$ as a constant.

Step4: Differentiate with respect to $t$

Using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we get $\frac{dy}{dt}=\frac{1}{x^{8}}-xt^{-2}=\frac{1}{x^{8}}-\frac{x}{t^{2}}$.

Answer:

$\frac{dy}{dx}=-\frac{8t}{x^{9}}+\frac{1}{t}$
$\frac{dy}{dt}=\frac{1}{x^{8}}-\frac{x}{t^{2}}$