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This is should be a straightforward question for me but I'm blacking out right now.

Let $g$ be a differentiable function that satisfies $g(x) + x^3 \sin(g(x)) = x^4 + 4x$ around $x=1$.

If $g(1) = \frac{\pi}{2}$ , find the value of $g'(1)$.

Now, am I plugging in $\frac{\pi}{2}$ in all the places where it says $g(x)$ in the equation? What is meant by the "around $x=1$" part; isn't this somewhat redundant information?

If the above statements are true, then I'm just isolating $g(x)$, taking the derivative, and plugging in $\frac{\pi}{2}$ as $g(x)$, correct?

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What are $x3$ and $x4$? –  littleO Nov 12 '12 at 8:02
    
Sorry, the powers didn't carry though. Corrected. –  EngGenie Nov 12 '12 at 8:03
    
“around $x=1$” just means that the formula holds for all $x$ in some open interval containing $1$. Further away, the equation may possibly not hold. –  Harald Hanche-Olsen Nov 12 '12 at 8:11

2 Answers 2

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I do not think you will succeed in isolating $g(x)$. (For sure, I would not be able to.)

Just differentiate immediately. We get $g'(x)+x^3g'(x)\cos(g(x))+3x^2\sin(g(x))=4x^3+4$.

Finally, put $x=1$, and solve for $g'(1)$.

You will need $\cos(g(1))$ and $\sin(g(1))$, but these are available from the given information.

Remark: One cannot find $g'(1)$ just from the information we have at $x=1$. We need to know that the given equation holds in a neighbourhood of $x=1$.

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This is an implicit differentiation question. You should apply "$\frac{d}{dx}$" to both sides of the equation, use your differentiation rules (like the product and chain rule), then after you're done plug in the value for $g(1)$ given. Then you can isolate $g^\prime(1)$.

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