How do you differentiate # (e^(2x) + 2x) ^0.5#?
using the chain rule :(twice)
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To differentiate (e^(2x) + 2x)^0.5, you can use the chain rule.

Let ( u = e^{2x} + 2x ).

Rewrite the expression as ( u^{0.5} ).

Apply the power rule: ( \frac{d}{du} u^{0.5} = \frac{1}{2}u^{0.5} ).

Apply the chain rule: ( \frac{d}{dx} (e^{2x} + 2x)^{0.5} = \frac{1}{2}(e^{2x} + 2x)^{0.5} \cdot \frac{d}{dx}(e^{2x} + 2x) ).

Differentiate ( e^{2x} + 2x ) with respect to ( x ): ( \frac{d}{dx}(e^{2x} + 2x) = 2e^{2x} + 2 ).

Substitute back in: ( \frac{d}{dx} (e^{2x} + 2x)^{0.5} = \frac{1}{2}(e^{2x} + 2x)^{0.5} \cdot (2e^{2x} + 2) ).

Simplify if necessary.
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To differentiate (e^(2x) + 2x) ^0.5, you can use the chain rule. The derivative of a function raised to a power can be calculated using the chain rule. Let's denote the given function as u(x) = (e^(2x) + 2x) and v(x) = 0.5. Then, the function can be rewritten as y = u(x) ^ v(x). Now, using the chain rule, the derivative dy/dx is given by dy/dx = v(x) * u(x)^(v(x)1) * u'(x) + ln(u(x)) * u(x) ^ v(x) * v'(x). Substituting the values of u(x) and v(x), and their derivatives, we get dy/dx = 0.5 * (e^(2x) + 2x)^(0.5) * (2e^(2x) + 2) + ln(e^(2x) + 2x) * (e^(2x) + 2x)^(0.5) * 0. Applying algebraic simplification, the final expression for the derivative dy/dx is dy/dx = (e^(2x) + 1) / (2(e^(2x) + 2x)^(0.5)).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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