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makvit [3.9K]
1 year ago
5

exercise 5.2.6. let g be defined on an interval a, and let c ∈ a. (a) explain why g′(c) in definition 5.2.1 could have been give

n by g′(c)
Mathematics
1 answer:
Dahasolnce [82]1 year ago
3 0

From definition 5.2.1:

g'(c) = limₓ→c  g(x) ₋ g(c)/x₋c

Let h = x ₋ c

Then x = c₊h

substituting in the above defination we have:

g'(c) = limc₊h→c  g(c₊h) ₋ g(c)/c₊h₋c

=  limc₊h→c g(c₊h) ₋ g(c)/h

x → c ⇒ x ₋ c → 0 ⇒ h → 0

Hence we have,

limc₊h→c  g(c₊h) ₋ g(c)/h

Therefore we have defined clearly why g′(c) in definition 5.2.1 could have been given by g′(c).

Learn more about limits and continuity here:

brainly.com/question/24637240

#SPJ4

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Help with 30 please. thanks.​
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Answer:

See Below.

Step-by-step explanation:

We have the equation:

\displaystyle  y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}

And we want to show that:

\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}

Instead of differentiating directly, we can first square both sides:

\displaystyle y^2 = 3e^{2x} -4x + 1

We can find the first derivative through implicit differentiation:

\displaystyle 2y \frac{dy}{dx}  = 6e^{2x} -4

Hence:

\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}

And we can find the second derivative by using the quotient rule:

\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}

Substitute:

\displaystyle y\left(\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\right) + \left(\frac{3e^{2x}-2}{y}\right)^2 =6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}- \left(3e^{2x} -2\right)^2}{y^2} + \frac{\left(3e^{2x}-2\right)^2}{y^2}= 6e^{2x}

Combine fractions:

\displaystyle \frac{\left(6y^2e^{2x}-\left(3e^{2x} - 2\right)^2\right) +\left(\left(3e^{2x}-2\right)^2\right)}{y^2} = 6e^{2x}

Simplify:

\displaystyle \frac{6y^2e^{2x}}{y^2} = 6e^{2x}

Simplify:

6e^{2x} \stackrel{\checkmark}{=} 6e^{2x}

Q.E.D.

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