# How do you differentiate 12sqrtx+12^(3sqrtx)+12^(4sqrtx)?

Aug 21, 2015

${y}^{'} = \frac{12 + 3 \cdot {12}^{3 \sqrt{x}} \cdot \ln \left(12\right) + 4 \cdot {12}^{4 \sqrt{x}} \cdot \ln \left(12\right)}{2 \sqrt{x}}$

#### Explanation:

All you really need to use in order to differentiate this function is the chain rule for a constant raised to a variable power.

Now, for a constant $a$ raised to avariable power $x$, you have

$\textcolor{b l u e}{{a}^{x} = {e}^{x \cdot \ln \left(a\right)}}$

This means that you can write, using th chain function for ${e}^{u}$, with $u = x \cdot \ln \left(a\right)$

$\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = \frac{d}{\mathrm{du}} \left({e}^{u}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {\underbrace{{e}^{x \cdot \ln \left(a\right)}}}_{\textcolor{red}{= {a}^{x}}} \cdot \frac{d}{\mathrm{dx}} \left(x \cdot \ln \left(a\right)\right)$

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {a}^{x} \cdot \ln \left(a\right)}$

Your function can be rewritten as

$y = 12 \cdot {x}^{\frac{1}{2}} + {12}^{3 \cdot {x}^{\frac{1}{2}}} + {12}^{4 \cdot {x}^{\frac{1}{2}}}$

The derivative of $y$ will be

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(12 {x}^{\frac{1}{2}}\right)\right] + \left[\frac{d}{\mathrm{dx}} {12}^{3 \cdot {x}^{\frac{1}{2}}}\right] + \frac{d}{\mathrm{dx}} \left({12}^{4 \cdot {x}^{\frac{1}{2}}}\right)$

As you can see, you're going to have to use the chain rule again for the last two derivatives. More specifically, you'll have ${12}^{u}$, with $u = 3 {x}^{\frac{1}{2}}$, and ${12}^{t}$, with $t = 4 {x}^{\frac{1}{2}}$.

This wil get you

$\frac{d}{\mathrm{dx}} \left({12}^{u}\right) = \frac{d}{\mathrm{du}} {12}^{u} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left({12}^{u}\right) = {12}^{u} \cdot \ln \left(12\right) \cdot \frac{d}{\mathrm{dx}} \left(3 \cdot {x}^{\frac{1}{2}}\right)$

$\frac{d}{\mathrm{dx}} \left({12}^{3 {x}^{\frac{1}{2}}}\right) = {12}^{3 {x}^{\frac{1}{2}}} \cdot \ln \left(12\right) \cdot \frac{3}{2} {x}^{- \frac{1}{2}}$

and

$\frac{d}{\mathrm{dx}} \left({12}^{t}\right) = \frac{d}{\mathrm{dt}} {12}^{t} \cdot \frac{d}{\mathrm{dx}} \left(t\right)$

$\frac{d}{\mathrm{dx}} \left({12}^{4 {x}^{\frac{1}{2}}}\right) = {12}^{4 {x}^{\frac{1}{2}}} \cdot \ln \left(12\right) \cdot 2 \cdot {x}^{- \frac{1}{2}}$

Plug this into your target derivative to get

${y}^{'} = 6 \cdot {x}^{- \frac{1}{2}} + {12}^{3 {x}^{\frac{1}{2}}} \cdot \ln \left(12\right) \cdot \frac{3}{2} {x}^{- \frac{1}{2}} + {12}^{4 {x}^{\frac{1}{2}}} \cdot \ln \left(12\right) \cdot 2 \cdot {x}^{- \frac{1}{2}}$

You can simplify this by using ${x}^{- \frac{1}{2}}$ as a common factor

${y}^{'} = {x}^{- \frac{1}{2}} \cdot \frac{1}{2} \cdot \left[12 + 3 \cdot {12}^{3 {x}^{\frac{1}{2}}} \cdot \ln \left(12\right) + 4 \cdot {12}^{4 {x}^{\frac{1}{2}}} \cdot \ln \left(12\right)\right]$

Finally, you can rewrite this as

${y}^{'} = \textcolor{g r e e n}{\frac{12 + 3 \cdot {12}^{3 \sqrt{x}} \cdot \ln \left(12\right) + 4 \cdot {12}^{4 \sqrt{x}} \cdot \ln \left(12\right)}{2 \sqrt{x}}}$

Aug 21, 2015

For $y = {12}^{\sqrt{x}} + {12}^{\sqrt[3]{x}} + {12}^{\sqrt[4]{x}}$, I get: $y ' = \ln \frac{12}{12} \left[{12}^{\sqrt{x}} \frac{6}{\sqrt{x}} + {12}^{\sqrt[3]{x}} \frac{4}{\sqrt[3]{x}} ^ 2 + {12}^{\sqrt[4]{x}} \frac{3}{\sqrt[4]{x}} ^ 3\right]$

#### Explanation:

I wonder if the intended question was to differentiate:

$y = {12}^{\sqrt{x}} + {12}^{\sqrt[3]{x}} + {12}^{\sqrt[4]{x}}$

Which uses the same ideas as the other answer, but the exponents are different:

$y = {12}^{\sqrt{x}} + {12}^{\sqrt[3]{x}} + {12}^{\sqrt[4]{x}} = {12}^{{x}^{\frac{1}{2}}} + {12}^{{x}^{\frac{1}{3}}} + {12}^{{x}^{\frac{1}{4}}}$

And the derivative will be:

$y ' = {12}^{{x}^{\frac{1}{2}}} \ln 12 \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}}\right) + {12}^{{x}^{\frac{1}{3}}} \ln 12 \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{3}}\right) + {12}^{{x}^{\frac{1}{4}}} \ln 12 \frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{4}}\right)$

$= {12}^{{x}^{\frac{1}{2}}} \left(\ln 12\right) \left[\frac{1}{2} {x}^{- \frac{1}{2}}\right] + {12}^{{x}^{\frac{1}{3}}} \left(\ln 12\right) \left[\frac{1}{3} {x}^{- \frac{2}{3}}\right] + {12}^{{x}^{\frac{1}{4}}} \left(\ln 12\right) \left[\frac{1}{4} {x}^{- \frac{3}{4}}\right]$

 = ln12/12 [ 12^(x^(1/2))6 x^(-1/2) + 12^(x^(1/3)) 4x^(-2/3) + 12^(x^(1/4)) 3 x^(-3/4)

$= \ln \frac{12}{12} \left[{12}^{\sqrt{x}} \frac{6}{\sqrt{x}} + {12}^{\sqrt[3]{x}} \frac{4}{\sqrt[3]{x}} ^ 2 + {12}^{\sqrt[4]{x}} \frac{3}{\sqrt[4]{x}} ^ 3\right]$