# Find the first partial derivatives of the following functions. f(x,y)=(3xy+4y^{2}+1)^{5}

Find the first partial derivatives of the following functions.
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{5}}}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

joshyoung05M
Step 1:Given
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{5}}}$$
Step 2:Solution
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{5}}}$$
$$\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={5}{\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{5}-{1}}}{\left({3}{y}\right)}={15}{y}{\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{4}}}{{f}_{{{y}}}{\left({x},{y}\right)}}$$
$$\displaystyle={5}{\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{5}-{1}}}{\left({3}{x}+{4}{\left({2}\right)}{y}^{{{2}-{1}}}\right)}$$
$$\displaystyle={5}{\left({3}{x}+{8}{y}\right)}{\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{4}}}$$
$$\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={15}{y}{\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{4}}}$$
$$\displaystyle{{f}_{{{y}}}{\left({x},{y}\right)}}={5}{\left({3}{x}+{8}{y}\right)}{\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{4}}}$$
###### Have a similar question?
content_user

Answer is given below (on video)