# How do you differentiate  f(x) = sqrt (3^x) / x^2?

##### 1 Answer
Jun 8, 2017

$f ' \left(x\right) = \frac{\sqrt{{3}^{x}} \left(x \ln 3 - 4\right)}{2 {x}^{3}} , \mathmr{and} ,$

$f ' \left(x\right) = \frac{\left(\sqrt{{3}^{x}}\right) \left(\ln \left({3}^{x}\right) - 4\right)}{2 {x}^{3}} .$

#### Explanation:

$f \left(x\right) = \frac{\sqrt{{3}^{x}}}{x} ^ 2 = {\left({3}^{x}\right)}^{\frac{1}{2}} / {x}^{2} = {\left({3}^{\frac{1}{2}}\right)}^{x} \cdot {x}^{-} 2 = {\left(\sqrt{3}\right)}^{x} \cdot {x}^{-} 2.$

Now, we know that,

$\left(1\right) : F \left(x\right) = G \left(x\right) H \left(x\right) \Rightarrow F ' \left(x\right) = G \left(x\right) H ' \left(x\right) + G ' \left(x\right) H \left(x\right) .$

$\left(2\right) : \left({a}^{x}\right) ' = {a}^{x} \ln a .$

$\left(3\right) : \left({x}^{n}\right) ' = n {x}^{n - 1} .$

$\therefore f ' \left(x\right) = {\left(\sqrt{3}\right)}^{x} \left({x}^{-} 2\right) ' + \left({x}^{-} 2\right) \left({\left(\sqrt{3}\right)}^{x}\right) ' .$

=((sqrt3)^x(-2x^(-2-1))+(x^-2){(sqrt3)^xlnsqrt3}

$= - 2 {x}^{-} 3 \sqrt{{3}^{x}} + \frac{\sqrt{{3}^{x}} \ln 3}{2 {x}^{2}}$

$= - \frac{2 \sqrt{{3}^{x}}}{x} ^ 3 + \frac{\sqrt{{3}^{x}} \ln 3}{2 {x}^{2}}$

$\Rightarrow f ' \left(x\right) = \frac{\sqrt{{3}^{x}} \left(x \ln 3 - 4\right)}{2 {x}^{3}} , \mathmr{and} ,$

$f ' \left(x\right) = \frac{\left(\sqrt{{3}^{x}}\right) \left(\ln \left({3}^{x}\right) - 4\right)}{2 {x}^{3}} .$

Enjoy Maths.!