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Sergeu [11.5K]
3 years ago
14

Simplify by expressing fractional exponents instead of radicals

Mathematics
1 answer:
igomit [66]3 years ago
3 0

Answer:

a^{\frac{1}{2}}b^{\frac{1}{2}}

Step-by-step explanation:

Given:

The expression in radical form is given as:

\sqrt{ab}

We need to express this in fractional exponent form.

We know that,

\sqrt a=a^{\frac{1}{2}}

Also, \sqrt{ab}=\sqrt a\times \sqrt b

Now, clubbing both the properties of square root function, we can rewrite the given expression as:

\sqrt{ab}=\sqrt a \times \sqrt b\\\\\sqrt{ab}=a^{\frac{1}{2}}\times b^{\frac{1}{2}}\\\\\therefore \sqrt{ab}=a^{\frac{1}{2}}b^{\frac{1}{2}}

So, the given expression in fractional exponents form is a^{\frac{1}{2}}b^{\frac{1}{2}}.

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Any help with an explanation would be appreciated!
Lilit [14]

Problem 1

We'll use the product rule to say

h(x) = f(x)*g(x)

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

Then plug in x = 2 and use the table to fill in the rest

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

h ' (2) = f ' (2)*g(2) + f(2)*g ' (2)

h ' (2) = 2*3 + 2*4

h ' (2) = 6 + 8

h ' (2) = 14

<h3>Answer: 14</h3>

============================================================

Problem 2

Now we'll use the quotient rule

h(x) = \frac{f(x)}{g(x)}\\\\h'(x) = \frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}\\\\h'(2) = \frac{f'(2)*g(2)-f(2)*g'(2)}{(g(2))^2}\\\\h'(2) = \frac{2*3-2*4}{(3)^2}\\\\h'(2) = \frac{6-8}{9}\\\\h'(2) = -\frac{2}{9}\\\\

<h3>Answer:  -2/9</h3>

============================================================

Problem 3

Use the chain rule

h(x) = f(g(x))\\\\h'(x) = f'(g(x))*g'(x)\\\\h'(2) = f'(g(2))*g'(2)\\\\h'(2) = f'(3)*g'(2)\\\\h'(2) = 3*4\\\\h'(2) = 12\\\\

<h3>Answer:  12</h3>
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