# Higher-Order Derivatives f(x)=2x^4-3x^3+2x^2+x+4 find f^10(x)=?

Higher-Order Derivatives
$f\left(x\right)=2{x}^{4}-3{x}^{3}+2{x}^{2}+x+4$
find ${f}^{10}\left(x\right)=?$
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Aubree Mcintyre
We will apply the power rule to find the derivatives.
We find ${f}^{1}\left(x\right)$ first
$f\left(x\right)=2{x}^{4}-3{x}^{3}+2{x}^{2}+x+4$
${f}^{1}\left(x\right)=2\left(4{x}^{3}\right)-3\left(3{x}^{2}\right)+2\left(2x\right)+1$
${f}^{1}\left(x\right)=8{x}^{3}-9{x}^{2}+4x+1$
Step 2
Then we differentiate ${f}^{1}\left(x\right)\to \ge t{f}^{2}\left(x\right)$
${f}^{1}\left(x\right)=8{x}^{3}-9{x}^{2}+4x+1$
${f}^{2}\left(x\right)=8\left(3{x}^{2}\right)-9\left(2x\right)+4$
${f}^{2}\left(x\right)=24{x}^{2}-18x+4$
Then we differentiate ${f}^{2}\left(x\right)\to \ge t{f}^{3}\left(x\right)$
${f}^{2}\left(x\right)=24{x}^{2}-18x+4$
${f}^{3}\left(x\right)=24\left(2x\right)-18\left(1\right)$
${f}^{3}\left(x\right)=48x-18$
Then we differentiate ${f}^{3}\left(x\right)\to \ge t{f}^{4}\left(x\right)$
${f}^{3}\left(x\right)=48x-18$
${f}^{4}\left(x\right)=48\left(1\right)$
${f}^{4}\left(x\right)=48$
Then we differentiate ${f}^{4}\left(x\right)\to \ge t{f}^{5}\left(x\right)$
${f}^{4}\left(x\right)=48$
${f}^{5}\left(x\right)=0$
Since the 5th order derivative is 0, so the other higher-order derivatives will be 0 too.
Result: ${f}^{10}\left(x\right)=0$