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3 years ago

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Find the derivative of the function by using the Product Rule. Simplify your answer. f(x) = (x - 3)(x + 3)

2 answers:

Minchanka [31]3 years ago

8 0

Answer:

$f'(x)=2x$

Step-by-step explanation:

To Find :Find the derivative of the function by using the Product Rule. Simplify your answer. f(x) = (x - 3)(x + 3)

Solution :

$f(x) = (x - 3)(x + 3)$

We will use chain rule of product

Formula : $uv=u \times v' +v \tyimes u'$

= $(x-3) \times 1+(x+3) \times 1$

= $(x-3)+(x+3)$

= $2x$

So, $f'(x)=2x$

Hence the derivative of the function by using the Product Rule is 2x

Morgarella [4.7K]3 years ago

3 0

Answer:

$f'(x)=2x$

Step-by-step explanation:

Given : Function $f(x)=(x-3)(x+3)$

To find : The derivative of the function by using the Product Rule ?

Solution :

The product rule of derivative is

$\frac{d}{dx}(u\cdot v)=uv'+vu'$

Here, u=x-3 and v=x+3

$\frac{d}{dx}((x-3)\cdot (x+3))=(x-3)\frac{d}{dx}(x+3)+(x+3)\frac{d}{dx}(x-3)$

$\frac{d}{dx}((x-3)\cdot (x+3))=(x-3)1+(x+3)1$

$\frac{d}{dx}((x-3)\cdot (x+3))=x-3+x+3$

$\frac{d}{dx}((x-3)\cdot (x+3))=2x$

Therefore, $f'(x)=2x$

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Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

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$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

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