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I have an equation:

$$y = \Bigl(x + f(x^2 - 1)\Bigr)^{1/2},$$ (so a square root function.) I am asked to find $dy/dx$ when $x = 3$, given that $f(8) = 0$, and $f'(8) = 3$.

I apply the chain rule as I usually do for this problem, derivative of the inside times derivative of the outside. I get:

$$1/2\Bigl(x + f(x^2-1)\Bigr)^{-1/2} \Bigl(1 + f'(x^2-1)\Bigr).$$ However, when my teacher did this example, she multiplied the $f'(x^2-1)$ part by $2x$. Why did she do so?

Thank you!

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    $\begingroup$ What if $f$ was, say $\sin(x^2+1)$, for example? $\endgroup$ – Moo Oct 19 '16 at 1:21
  • $\begingroup$ Because you have to apply the chain rule a second time to find $\frac d{dx}f(x^2-1)$. $\endgroup$ – amd Oct 19 '16 at 7:41
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Hint: Rewrite $f(x^2-1)$ as $(f\circ g )(x)$ where $g(x)=x^2-1$, and then take the derivative of the composition using the chain rule.

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Consider $$y = \Bigl(x + f[t(x)]\Bigr)^{1/2}$$ An easy way could be logarithmic differentiation $$y = \Bigl(x + f[t(x)]\Bigr)^{1/2}\implies \log(y)=\frac 12 \log\Bigl(x + f[t(x)]\Bigr)$$ $$\frac {y'}y=\frac 12 \frac{1+\frac{df}{dt}\times \frac{dt}{dx}}{x + f[t(x)]}\implies y'= \frac{1+\frac{df}{dt}\times \frac{dt}{dx}}{2\Bigl(x + f[t(x)]\Bigr)^{1/2}}$$

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