Question

# Find all derivatives of the function y=\frac{1}{2x-3}

Derivatives
Find all derivatives of the function $$\displaystyle{y}={\frac{{{1}}}{{{2}{x}-{3}}}}$$

2021-06-08
Step 1
We find first few derivatives using the power rule and chain rule
$$\displaystyle{y}={\frac{{{1}}}{{{2}{x}-{3}}}}={\left({2}{x}-{3}\right)}^{{-{1}}}$$
$$\displaystyle{y}^{{{\left({1}\right)}}}=-{1}{\left({2}{x}-{3}\right)}^{{-{2}}}{\left({2}\right)}=-{2}{\left({2}{x}-{3}\right)}^{{-{2}}}={\left(-{1}\right)}{\left({2}\right)}{\left({1}!\right)}{\left({2}{x}-{3}\right)}^{{-{2}}}$$
$$\displaystyle{y}^{{{\left({2}\right)}}}={4}{\left({2}{x}-{3}\right)}^{{-{3}}}{\left({2}\right)}={8}{\left({2}{x}-{3}\right)}^{{-{3}}}={\left(-{1}\right)}^{{{2}}}{\left({2}\right)}{\left({2}!\right)}{\left({2}{x}-{3}\right)}^{{-{3}}}$$
$$\displaystyle{y}^{{{\left({3}\right)}}}=-{12}{\left({2}{x}-{3}\right)}^{{-{3}}}{\left({2}\right)}=-{24}{\left({2}{x}-{3}\right)}^{{-{4}}}={\left(-{1}\right)}^{{{3}}}{\left({2}\right)}{\left({3}!\right)}{\left({2}{x}-{3}\right)}^{{-{4}}}$$
$$\displaystyle{y}^{{{\left({4}\right)}}}={36}{\left({2}{x}-{3}\right)}^{{-{3}}}{\left({2}\right)}={72}{\left({2}{x}-{3}\right)}^{{-{5}}}={\left(-{1}\right)}^{{{4}}}{\left({2}\right)}{\left({4}!\right)}{\left({2}{x}-{3}\right)}^{{-{5}}}$$
Step 2
Then we find a pattern for integer n
$$\displaystyle{y}^{{{\left({n}\right)}}}={\left(-{1}\right)}^{{{n}}}{\left({2}\right)}{\left({n}!\right)}{\left({2}{x}-{3}\right)}^{{-{n}-{1}}}$$
Answer: $$\displaystyle{y}^{{{\left({n}\right)}}}={\left(-{1}\right)}^{{{n}}}{\left({2}\right)}{\left({n}!\right)}{\left({2}{x}-{3}\right)}^{{-{n}-{1}}}$$