Find the derivatives y=(1-4x+7x^{5})^{30}

chillywilly12a 2021-10-03 Answered
Find the derivatives
\(\displaystyle{y}={\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{30}}}\)

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Expert Answer

dessinemoie
Answered 2021-10-04 Author has 12577 answers
Step 1 Background concept
If a function is given,
y=f(ax+b)
then it's derivative will be given as,
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{f{{\left({a}{x}+{b}\right)}}}\)
\(\displaystyle={f}'{\left({a}{x}+{b}\right)}\cdot{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({a}{x}+{b}\right)}\)
Here, It is given as,
\(\displaystyle{y}={\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{30}}}\)
Step 2 Derivative of the given function
Therefore, the derivative of the given function will be,
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{30}}}\)
\(\displaystyle\Rightarrow{f}'{\left({x}\right)}={30}{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{29}}}\cdot{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}\)
\(\displaystyle\Rightarrow{f}'{\left({x}\right)}={30}{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{29}}}\cdot{\left(-{4}+{35}{x}^{{{4}}}\right)}\cdot{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{x}\)
\(\displaystyle\Rightarrow{f}'{\left({x}\right)}={30}{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{29}}}\cdot{\left(-{4}+{35}{x}^{{{4}}}\right)}\)
\(\displaystyle\Rightarrow{f}'{\left({x}\right)}={\left(-{120}+{1050}{x}^{{{4}}}\right)}\cdot{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{29}}}\)
So, derivative of the given function, \(\displaystyle={\left(-{120}+{1050}{x}^{{{4}}}\right)}\cdot{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{29}}}\).
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