Question
Find $f^{\prime}(2)$ if $f(x)=\frac{4 x^{5}+3 x^{3}+2 x^{2}+24}{x^{2}}$
✓
Answer
$f(x)=\frac{4 x^5+3 x^3+2 x^2+24}{x^2}$
$ =\frac{4 x^5}{x^2}+\frac{3 x^3}{x^2}+\frac{2 x^2}{x^2}+\frac{24}{x^2}$
$ =4 x^3+3 x+2+24 x^{-2}$
$ \therefore f^{\prime}(x)=4\left(3 x^2\right)+3(1)+0+24(-2) x^{-2-1}$
$ =12 x^2+3-48 x^{-3}$
$ =12 x^2+3-\frac{48}{x^3}$
$ \text { Putting, } x=2$
$ f^{\prime}(2)=12(2)^2+3-\frac{48}{(2)^3}$
$ =(12 \times 4)+3-\frac{48}{8}$
$ =48+3-6$
$ =45$
$ \text { Hence, if } f(x)=\frac{4 x^5+3 x^3+2 x^2+24}{x^2}$
$ \text { then } f^{\prime}(2)=45$
$ =\frac{4 x^5}{x^2}+\frac{3 x^3}{x^2}+\frac{2 x^2}{x^2}+\frac{24}{x^2}$
$ =4 x^3+3 x+2+24 x^{-2}$
$ \therefore f^{\prime}(x)=4\left(3 x^2\right)+3(1)+0+24(-2) x^{-2-1}$
$ =12 x^2+3-48 x^{-3}$
$ =12 x^2+3-\frac{48}{x^3}$
$ \text { Putting, } x=2$
$ f^{\prime}(2)=12(2)^2+3-\frac{48}{(2)^3}$
$ =(12 \times 4)+3-\frac{48}{8}$
$ =48+3-6$
$ =45$
$ \text { Hence, if } f(x)=\frac{4 x^5+3 x^3+2 x^2+24}{x^2}$
$ \text { then } f^{\prime}(2)=45$
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