Determine f(n)(x) if f(x)=1(2−x)
Rewrite f(x) to f(x)=(2−x)−1, then solve for the 1st derivative.
We have,
f′(x)=ddx(2−x)−1f′(x)=−1(2−x)−2ddx(2−x)f′(x)=−1(2−x)−2(−1)f′(x)=(2−x)−2
Solving for the 2nd derivative
f″(x)=ddx(2−x)−2f″(x)=−2(2−x)−3ddx(2−x)f″(x)=−2(2−x)−3(−1)f″(x)=2(2−x)−3
Solving for the 3rd derivative
f‴(x)=2ddx(2−x)−3f‴(x)=(2)(−3)(2−x)−4ddx(2−x)f‴(x)=−6(2−x)−4(−1)f‴(x)=6(2−x)−4
By solving the first, second and third derivative of the function. We get the pattern,
f(n)=n!(2−x)(−1−n)
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