How do you differentiate #root3(x)+root4(x)#?

Question

Charles Anctil · Accepted Answer

#=1/(3x^(2/3))+1/(4x^(3/4))#.

We know that, #d/dx(x^n)=n*x^(n-1)#, and, #d/dx(u+v)=(du)/dx+(dv)/dx# Therefore, #d/dx(root3x+root4x)=d/dxroot3x+d/dxroot4x# #=d/dxx^(1/3)+d/dxx^(1/4)# #=1/3*x^(1/3-1)+1/4*x^(1/4-1)# #=1/(3x^(2/3))+1/(4x^(3/4))#.

Elijah Abram · Answer

undefined To differentiate $ \sqrt{3}(x) + \sqrt{4}(x) $, we need to use the power rule of differentiation.

The power rule states that the derivative of $ x^n $ with respect to $ x $ is $ nx^{n-1} $, where $ n $ is a constant.

For $ \sqrt{3}(x) $, we can rewrite it as $ 3^{1/2}x $, and for $ \sqrt{4}(x) $, we can rewrite it as $ 4^{1/2}x $.

Therefore, the derivative of $ \sqrt{3}(x) + \sqrt{4}(x) $ is:

$$ \frac{d}{dx} \left( \sqrt{3}(x) + \sqrt{4}(x) \right) = \frac{d}{dx} \left( 3^{1/2}x + 4^{1/2}x \right) $$

$$ = \frac{d}{dx} \left( 3^{1/2}x \right) + \frac{d}{dx} \left( 4^{1/2}x \right) $$

$$ = \frac{1}{2} \cdot 3^{1/2}x^{(1/2) - 1} + \frac{1}{2} \cdot 4^{1/2}x^{(1/2) - 1} $$

$$ = \frac{1}{2} \cdot 3^{1/2}x^{-1/2} + \frac{1}{2} \cdot 4^{1/2}x^{-1/2} $$

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

$$ = \frac{1}{2} \cdot \frac{\sqrt{3}}{\sqrt{x}} + \frac{1}{2} \cdot \frac{\sqrt{4}}{\sqrt{x}} $$

$$ = \frac{1}{2} \cdot \frac{\sqrt{3}}{\sqrt{x}} + \frac{1}{2} \cdot \frac{2}{\sqrt{x}} $$

$$ = \frac{1}{2} \cdot \left( \frac{\sqrt{3} + 2}{\sqrt{x}} \right) $$

So, the derivative of $ \sqrt{3}(x) + \sqrt{4}(x) $ is $ \frac{1}{2} \cdot \left( \frac{\sqrt{3} + 2}{\sqrt{x}} \right) $.