Diff Example with Square Root

[Jason76]

Product Rule:

Given: \(\displaystyle f(x) g(x)\)

\(\displaystyle g(x)[f'(x)] + f(x)[g'(x)]\)

\(\displaystyle f(x) = 7\sqrt{x} \sin(x)\)

\(\displaystyle f(x) = 7 x^{1/2} \sin(x)\)

\(\displaystyle f'(x) = \sin(x) [\dfrac{d}{dx} 7 x^{1/2}] + 7x^{1/2} [\dfrac{d}{dx} \sin(x)]\)

\(\displaystyle f'(x) = \sin(x) \dfrac{7}{2}x^{-1/2} + 7x^{1/2} \cos(x)\)

 

[fcabanski]

The solution is correct. You can simplify it by finding the LCD, then adding like fractions. You can check derivatives at wolframalpha.com.

 

[Jason76]

Product Rule:

Given: \(\displaystyle f(x) g(x)\)

\(\displaystyle g(x)[f'(x)] + f(x)[g'(x)]\)

\(\displaystyle f(x) = 7\sqrt{x} \sin(x)\)

\(\displaystyle f(x) = 7(x^{1/2}) \sin(x)\)

\(\displaystyle f'(x) = \dfrac{d}{dx} 7(x^{1/2}) \sin(x)\)

\(\displaystyle f'(x) = \sin(x) \dfrac{7}{2}x^{-1/2} + 7x^{1/2} \cos(x)\)

\(\displaystyle f'(x) = \sin(x) \dfrac{7}{2 \sqrt{x}} + 7x^{1/2} \cos(x)\) - :smile: Can be simplified further, and homework may demand it.