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

Find the derivatives
$$\displaystyle{y}={\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{30}}}$$

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dessinemoie
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, $$\displaystyle={\left(-{120}+{1050}{x}^{{{4}}}\right)}\cdot{\left({1}-{4}{x}+{7}{x}^{{{5}}}\right)}^{{{29}}}$$.