How do you find the derivative of #5e^x sqrt(x)#?

Answer 1

I would use the Product Rule where if you have:

#y=f(x)g(x)# then #y'=f'(x)g(x)+f(x)g'(x)#

In your case you get: #y'=5e^xsqrt(x)+5e^x1/(2sqrt(x))# #=5e^x[sqrt(x)+1/(2sqrt(x))]=# #=5e^x[(2x+1)/(2sqrt(x))]#

By the way: It is a good idea, after you memorize the first derivative rules, to memorize: #d/(dx)(sqrtx)=1/(2sqrtx)#.

The square root function comes up a lot, because it is the length of a diagonal (hypotenuse of a right triagle). So it is involved in distance.

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Answer 2

Elijah Abram

To find the derivative of (5e^x \sqrt{x}), you can use the product rule of differentiation.

Let (u = 5e^x) and (v = \sqrt{x}).

Then, differentiate (u) with respect to (x) to get (u' = 5e^x), and differentiate (v) with respect to (x) to get (v' = \frac{1}{2\sqrt{x}}).

Apply the product rule: ((uv)' = u'v + uv').

Substitute the values of (u), (u'), (v), and (v') into the formula: ((5e^x \cdot \sqrt{x})' = (5e^x)(\sqrt{x}) + (5e^x)(\frac{1}{2\sqrt{x}})).

Simplify the expression to get the derivative: ((5e^x \cdot \sqrt{x})' = 5e^x\sqrt{x} + \frac{5e^x}{2\sqrt{x}}).

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Answer from HIX Tutor

When evaluating a one-sided 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.