Question

Find the derivative of $f\left(x\right)=1+x+x^2+x^3+\cdots +x^{50}$ at $x=1.$

Answer 1

Given : $f\left(x\right)=1+x+x^2+x^3+\cdots +x^{50}$
Differentiation with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=0+1\cdot x^{1-1}+2\cdot x^{2-1}+3\cdot x^{3-1}+\cdots +50\cdot x^{50-1}$
$\Rightarrow f^{\prime }\left(x\right)=1+2x+3x^2+\cdots +50x^{49}$
Putting $x=1$ in $f^{\prime }\left(x\right)$
$\Rightarrow f^{\prime }\left(1\right)=1+2\left(1\right)+3(1{)}^2+\cdots +50(1{)}^{49}$
$\Rightarrow f^{\prime }\left(1\right)=1+2+3+\cdots +50$
$\Rightarrow f^{\prime }\left(1\right)=\frac{50(50+1)}{2}$
$\Rightarrow f^{\prime }\left(1\right)=25\times 51$
$\Rightarrow f^{\prime }\left(1\right)=1275$
Hence, $f^{\prime }\left(x\right)$ at $x=1$ is $1275$.