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# Write the composite function in the form $f(g(x)).$ [Identify the inner function $u = g(x)$ and the outer function $y = f(u).$ ] Then find the derivative $dy/ dx.$$y = \sqrt{2 - e^x}$

## $\frac{d y}{d x}=-\frac{e^{x}}{2 \sqrt{2-e^{x}}}$

Derivatives

Differentiation

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here we have a composite function and we're going to start by identifying the inside function and outside function. So you're square root Symbol is serving as a grouping symbol, and inside it we find our inside function G of X. That would be to minus E to the X, and so the outside function would be the square root of that. So the square root of X and for the sake of differentiation, I like to call square root X X to the 1/2 power. That way I can use the power rule. OK, now it's time to find the derivative D Y d. X. So according to the chain rule, we start by finding the derivative of the outside function. And I'm going to rewrite the function as why equals two minus e to the X to the 1/2 power. So the derivative of the outside function would be using the power rule On the 1/2 power function, we bring down the 1/2 and then we take the inside to the negative 1/2. We have to subtract one from the old power to get the new power. Now we multiply by the derivative of the inside and the derivative of To minus each of the X would be negative e to the X. Okay, so we have our derivative, and now it's a matter of just simplifying. So what we want to do is write this as a single fraction, and we can get rid of our negative exponents by bringing that factor to the denominator. So we have negative E to the X over two times two minus each of the X to the 1/2 power. And then finally, instead of writing that quantity to the 1/2 power, we can change it back into a radical, which is kind of nice, because if your problem was given as a radical, it's nice to give your answer in that formas well. So D Y DX is negative each of the X over two times the square root of two minus E to the X. And it doesn't matter if you have your negative in the numerator or the denominator, or out in front of the fraction it's equivalent

Oregon State University

Derivatives

Differentiation

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