Trig Derivative with Square Root

[Jason76]

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

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

\(\displaystyle f'(x) = [\sin(x)][\dfrac{d}{dx}(7x^{1/2})] + [7x^{2}][\dfrac{d}{dx}(\sin(x))]\) - product rule: \(\displaystyle [g][f'] + [f][g']\) given \(\displaystyle (f)(g)\)

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

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

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

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

\(\displaystyle f'(x) = [\sin(x)][\dfrac{14x}{2})^{-1/2}] + [7x^{1/2}][\cos(x)]\)

\(\displaystyle f'(x) = [\sin(x)][7x^{-1/2}] + [7x^{1/2}][\cos(x)]\) Can this be simplified further?

 

[Deleted member 4993]

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

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

\(\displaystyle f'(x) = [\sin(x)][\dfrac{d}{dx}(7x^{1/2})] + [\dfrac{d}{dx}(\sin(x))]\) - product rule: \(\displaystyle [g][f'] + [f][g']\) given \(\displaystyle (f)(g)\)

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

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

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

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

Keep the multiplicative constant out while trying to differentiate.

 

[Jason76]

Keep the multiplicative constant out while trying to differentiate.

Please explain

 

[Deleted member 4993]

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

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

\(\displaystyle f'(x) = [\sin(x)][\dfrac{d}{dx}(7x^{1/2})] + [7x^{2}][\dfrac{d}{dx}(\sin(x))]\) - product rule: \(\displaystyle [g][f'] + [f][g']\) given \(\displaystyle (f)(g)\)

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

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

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

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

\(\displaystyle f'(x) = [\sin(x)][\dfrac{14x}{2})^{-1/2}] + [7x^{1/2}][\cos(x)]\)

\(\displaystyle f'(x) = [\sin(x)][7x^{-1/2}] + [7x^{1/2}][\cos(x)]\) Can this be simplified further?

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

\(\displaystyle \displaystyle f'(x) = 7*\left [\frac{1}{2}x^{-\frac{1}{2}}*\sin(x) \ + \ x^{\frac{1}{2}}*\cos(x)\right ]\) ..... according to me - this is simple enough

But you can further manipulate it to:

\(\displaystyle \displaystyle f'(x) = \dfrac{7*cos(x)}{2*\sqrt{x}}\left [2*x \ + \ \tan(x) \right ]\)