Carla Murphy

2021-12-21

If $f\left(2\right)=10$ and ${f}^{\prime }\left(x\right)={x}^{2}f\left(x\right)$ for all x, find $f{}^{″}\left(2\right)$

enhebrevz

Expert

Step 1
It is given that
${f}^{\prime }\left(x\right)={x}^{2}f\left(x\right)$
Substitute $x=2$
${f}^{\prime }\left(2\right)={2}^{2}×f\left(2\right)$
Substitute $f\left(2\right)=10$
${f}^{\prime }\left(2\right)=4×10=40$
${f}^{\prime }\left(x\right)={x}^{2}f\left(x\right)$
Differentiate using the product rule
$f{}^{″}\left(x\right)={\left({x}^{2}\right)}^{\prime }×f\left(x\right)+{x}^{2}×{f}^{\prime }\left(x\right)$
$f{}^{″}\left(x\right)=\left(2{x}^{2-1}\right)×f\left(x\right)+{x}^{2}×{f}^{\prime }\left(x\right)$
Substitute $x=2$
$f{}^{″}\left(2\right)=2×2×f\left(2\right)+{2}^{2}×{f}^{\prime }\left(2\right)$
Substitute $f\left(2\right)=10$ and ${f}^{\prime }\left(2\right)=40$
$f{}^{″}\left(2\right)=4×10+4×40=40+160=200$

Bertha Jordan

Expert

Step 1
Metod 1:
Divide by $f\left(x\right)$
$\frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}={x}^{2}$
Step 2
Differentiate both sides
$\frac{f{}^{″}\left(x\right)f\left(x\right)-{f}^{\prime }{\left(x\right)}^{2}}{{f\left(x\right)}^{2}}=2x$
Step 3
At $x=2$
${f}^{\prime }\left(2\right)={2}^{2}f\left(2\right)$
${f}^{\prime }\left(2\right)=4×10=40$
$\frac{f{}^{″}\left(2\right)f\left(2\right)-{f}^{\prime }{\left(2\right)}^{2}}{{f\left(2\right)}^{2}}=2×2$
$\frac{10f{}^{″}\left(2\right)-{40}^{2}}{{10}^{2}}=4$
$10f{}^{″}\left(2\right)=400+1600$
Step 4
Solve for $f{}^{″}\left(x\right)$
$f{}^{″}\left(2\right)=\frac{2000}{10}=200$
Step 5
Method 2 (Rose Winter):
Use the product rule
$f{}^{″}\left(x\right)=2xf\left(x\right)+{x}^{2}{f}^{\prime }\left(x\right)$
Step 6
As before, at $x=2$ we can calculate:
${f}^{\prime }\left(2\right)={2}^{2}f\left(2\right)$
${f}^{\prime }\left(2\right)=4×10=40$
Step 7
Substituting the values we get:
$f{}^{″}\left(2\right)=2×2×10+{2}^{2}×40=$
$40+160=$
$=200$

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