# Find the derivative using the appropriate rule or combination of rules.

Find the derivative using the appropriate rule or combination of rules.
$$\displaystyle{y}={\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}}}$$

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Ayesha Gomez
Step 1
We have to find the derivative of function:
$$\displaystyle{y}={\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}}}$$
We know the formula of derivatives,
$$\displaystyle{\frac{{{\left.{d}{x}\right.}^{{{n}}}}}{{{\left.{d}{x}\right.}}}}={n}{x}^{{{n}-{1}}}$$
$$\displaystyle{\frac{{{d}{a}{x}^{{{n}}}}}{{{\left.{d}{x}\right.}}}}={a}{\frac{{{\left.{d}{x}\right.}^{{{n}}}}}{{{\left.{d}{x}\right.}}}}$$ (where a is constant)
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{x}\right.}}}}={1}$$
$$\displaystyle{\frac{{{d}{a}}}{{{\left.{d}{x}\right.}}}}={0}$$
$$\displaystyle{\frac{{{d}{\left({f{{\left({x}\right)}}}\right)}^{{{n}}}}}{{{\left.{d}{x}\right.}}}}={n}{\left({f{{\left({x}\right)}}}\right)}^{{{n}-{1}}}{\frac{{{d}{f{{\left({x}\right)}}}}}{{{\left.{d}{x}\right.}}}}$$
Step 2
Differentiating the given function with respect to 't', we get
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{{d}{\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}{{{\left.{d}{t}\right.}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}-{1}}}{\frac{{{d}{\left({4}{t}+{9}\right)}}}{{{\left.{d}{t}\right.}}}}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left({4}{t}+{9}\right)}^{{{\frac{{{1}-{2}}}{{{2}}}}}}{\left({4}{\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{d}{9}}}{{{\left.{d}{t}\right.}}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left({4}{t}+{9}\right)}^{{-{\frac{{{1}}}{{{2}}}}}}{\left({4}\times{1}+{0}\right)}$$
$$\displaystyle={\frac{{{4}}}{{{2}}}}\times{\frac{{{1}}}{{{\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}}$$
$$\displaystyle={\frac{{{2}}}{{{\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}}$$
Hence, derivative of the function is $$\displaystyle{\frac{{{2}}}{{{\left({4}{t}+{9}\right)}^{{{\frac{{{1}}}{{{2}}}}}}}}}$$.