Answer:
Required relation is [tex]dy=\frac{-12dx}{x^{13}}[/tex].
Step-by-step explanation:
Given,
[tex]f(x)=\frac{1}{x^{12}}[/tex] be such that, [tex]dy=f'(x)dx[/tex].
To show relationship between a small change in x and corresponding change in y, consider,
[tex]dy=f'(x)dx[/tex]
[tex]=\frac{d}{dx}(f(x))dx[/tex]
[tex]=\frac{d}{dx}(x^{-12})dx[/tex]
[tex]=-12x^{-12-1}dx[/tex]
[tex]\therefore dy=\frac{-12dx}{x^{13}}[/tex]
which is the required relation.