Question

If g(x)+x sin(g(x)) =x^2 and g(1)=0, find g′(1).

g′(1)=

I keep getting 0 and I dont know where I'm going wrong. Please show step by step

Answer 1

We have
\[g(x)+x\sin (g(x))=x^2\]
Let's differentiate this
\[g'(x)+x\frac{d}{dx}(\sin g(x))+\sin g(x)\times \frac{d}{dx} x=\frac{d}{dx} (x^2)\]
Simplifying now
\[g'(x)+x\cos g(x)\times g'(x)+\sin g(x)=2x\]
Do you get till here?

Answer 2

why is it sin g(x) + sin g(x)? where did the second sin g(x) come from?

Answer 3

ive got that far minus the extra sin g(x)

Answer 4

It's because we have two functions x and sin (g(x))
\[\frac{d}{dx} p(x) q(x)\]
I've applied the multiplicative rule of differentiation there
\[\frac{d}{dx} p(x) q(x)=p'(x)q(x)+p(x)q'(x)\]

Answer 5

oo ok so you dont have to use chain rule on the sin g(x) then. Is that because it's a single variable inside the sin? if it was sin g(2x) would you need to use the chain rule?

Answer 6

We have to use chain rule, see the second term.
Cos (g(x) * g'(x)

Answer 7

right I hadnt looked at that part yet. So u did multiplication rule first then chain rule. Does it matter what order you do them in?

Answer 8

Yes, first the differentiation of main function and then multiplication with the inner function's derivative

Answer 9

ooo ok, thank you for your help!! :)

Answer 10

No problem :)