The Rivative

Remember the rules , soldiers. I will not post them here.

Back to the problem, $f(x) = \dfrac{2^x}{x^2}$ . This means

$f'(x) = \dfrac{\ln{2} \cdot 2^x \cdot x^2 - 2^x \cdot 2x}{x^4} \Rightarrow f'(x) = \dfrac{2^x ( x \ln{2} - 2)}{x^3}$ .

Applying another derivative, we get

$f'(x) = \dfrac{2^x ( x \ln{2} - 2)}{x^3} \Rightarrow f''(x) = \dfrac{(\ln{2} \cdot 2^x (x \ln{2} - 2)) x^3 - (2^x ( x \ln{2} - 2)) 3x^2 }{x^6}$ .

$\Rightarrow f''(x) = \dfrac{2^x (x^2 \: \ln^2{4} - 4 x \ln{2} + 6)}{x^4} \Rightarrow f''(2) = \dfrac{(2 \: \ln^2{4} - 4 \ln{2} + 3)}{2}$ .

Considering $\ln{2} \approx 0.69$ , we get that $f''(2) \approx 0.596$ . Thus, the biggest integer that is smaller than $59.6$ is $\boxed{59.}$

Let $y=\frac{2^{x}}{x^{2}}$

Now differentiating using division rule,

$y'=\frac{x^{2}*2^{x}ln(2)-2^{x}*2x}{x^{4}}$

You can express this as

$y'=yln(2)-\frac{2y}{x}$

Again differentiating,

$y''=y'ln(2)-\frac{x(2y')-2y}{x^{2}}$

Therefore,

$y''=y'ln(2)-\frac{2y'}{x}+\frac{2y}{x^{2}}$

Now, putting x=2;

$\boxed{y(2)=1}$

$\boxed{y'(2)=ln(2)-1}$

$y''(2)=ln(2)(ln(2)-1)-\frac{2(ln(2)-1)}{2}+\frac{1}{2}$

$=(ln2)^{2}-2ln(2)+\frac{3}{2}$

$=\boxed{0.5942}$ ...approx

Therefore, $100y''(2)=59.42$

Since answer has to be an integer, $100y''(2)=59$

Applying the quotient rule, {f^'}({x})=\frac{{x^2}\times{2^x}{ln2}-{2^x}\times{2x}}{x^4} {f^''}({x})=\frac{{x^4}\times(({2x}\times{2^x}({ln2})^2+{2^x}({ln2})\times{2x})-({2^x}\times{2}+{2x}\times{2^x}{ln2}))-({x^2}{2^x}{ln2}-{2^x}{2x})\times{4x^3}}{x^8} Therefore {100}{f^''}{(2)}=\boxed{59}

f(x) = $2^x/x^2$

Thus, f'(x)= $d/dx(2^x/x^2)$

Use the product rule, $d/dx(u*v)$ = $v ( du)/( dx)+u ( dv)/( dx)$ , where $u = 2^x$ and $v = 1/x^2$ :

f'(x)= $(d/dx(2^x))/x^2+2^x (d/dx(1/x^2))$

The derivative of $2^x$ is $2^x log(2)$ :

f'(x)= $2^x (d/dx(1/x^2))+2^x log(2)/x^2$

Use the power rule, $d/dx(x^n)$ = $n x^(n-1)$ , where n = -2:-------> $d/dx(1/x^2) = d/dx(x^(-2))$ = $-2 x^(-3)$ :

f ' (x) = $(2^x log(2))/x^2+(-2)/x^3 2^x$

Differentiating it again

f "(x) = $d/dx((2^x (x log(2)-2))/x^3)$ = $(2^x (x^2 log^2(2)-4 x log(2)+6))/x^4$

Therefore, f" (2) = $3/2+log^2(2)-log(4)$ that's approximately 0.594.

Hence 100*f"(2) = 59.4

taking ln for both sides ln(y) = ln (2)^x - ln (x)^2

ln(y) =x ln (2) - 2 ln (x)

y'/y= ln (2) - 2/x

y' = y ( ln(2)- 2/x )

y'' = y' ( ln(2) - 2/x) + 2y/x^2

y'' = y [ (ln (2) - 2/x)^2 + 2/x^2 ]

when x = 2 ... y=1

so y'' = [ ln(2)- 1 ] ^2 + 0.5

100y'' = 100*0.5941

so 100y'' = 59.41