Question

What is the derivative of: y= sqrt(x)sin(x^2)

Answer 1

\[y=\sqrt{x}\sin \left( x^2 \right)\]

Answer 2

Use product rule and chain rule.

Answer 3

I get \[\sqrt{x}2xcos \left( x^2 \right)+\frac{ sinx^2 }{ 2\sqrt{x} }\] , just seems like a strange answer?

Answer 4

hmm...

Answer 5

what're you hmming about?

Answer 6

Just that I feel lazy at the moment.\[\rm{d \over dx}(f(x)g(x)) =f'g +fg' \]and\[\rm {d \over dx}f(g(x)) = g'(x) f'(g(x))\]

Answer 7

g(x) = x^(1/2)
g'(x) = \[\frac{1}{2(x)^{\frac{1}{2}}}\]


for
s(x) = sin(x^2)

m(x) = sin(x)
m'(x) = cos(x)

d(x) = x^(2)
d'(x) = 2x

use chain rule

m'( d(x) )*(d'(x))

so we have

s'(x) = cos(x^2)*(2x)

so finally we just apply product rule using the following

g(x) = x^(1/2)
g'(x) = \[\frac{1}{2(x)^{\frac{1}{2}}}\]

s(x) = sin(x^2)
s'(x) = cos(x^2)*(2x)


product rule is

g'(x)s(x) + s'(x)g(x) = f'(x)

Answer 8

so we get

\[(\frac{1}{2\sqrt{x}})\sin(x^2) + \cos(x^2)2x \sqrt{x}\]

Answer 9

Ok, great. That was the answer I had. Just seemed like a strange derivative. Thought I had made a mistake somewhere. Thanks a bunch! :)

Answer 10

To avoid problems with derivatives split each part of the problem up into easier functions and solve the derivative then just memorize the product, quient and chain rule and you wont ever make a mistake

Answer 11

also use wolframalpha.com to check your answers, you shouldnt have another problem with derivatives if you use my method

Answer 12

https://www.wolframalpha.com/input/?i=d%2Fdx+%28x%29^%281%2F2%29sin%28x^2%29

Answer 13

Great, I'll keep that in mind. It was just the simplifying that worried me. I find it hard to tell when I can't simplify any more and when I should go further.