MARKING BRAINLEIST BUT PLS BE QUICK

Mathematics

2 answers:

uysha [10]3 years ago

8 0

It’s d! Or either b I would choose d !

Leno4ka [110]3 years ago

8 0

Answer:

A. $\frac{81m^{4} }{n{4} }$

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For the composite function, identify an inside function and an outside function and write the derivative with respect to x of th

alexira [117]

Answer:

The inner function is $h(x)=4x^2 + 8$ and the outer function is $g(x)=3x^5$ .

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

Step-by-step explanation:

A composite function can be written as $g(h(x))$ , where $h$ and $g$ are basic functions.

For the function $f(x)=3(4x^2+8)^5$ .

The inner function is the part we evaluate first. Frequently, we can identify the correct expression because it will appear within a grouping symbol one or more times in our composed function.

Here, we have $4x^2+8$ inside parentheses. So $h(x)=4x^2 + 8$ is the inner function and the outer function is $g(x)=3x^5$ .

The chain rule says:

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)$

It tells us how to differentiate composite functions.

The function $f(x)=3(4x^2+8)^5$ is the composition, $g(h(x))$ , of

outside function: $g(x)=3x^5$

inside function: $h(x)=4x^2 + 8$

The derivative of this is computed as

$\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=3\frac{d}{dx}\left(\left(4x^2+8\right)^5\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=u^5,\:\:u=\left(4x^2+8\right)\\\\3\frac{d}{du}\left(u^5\right)\frac{d}{dx}\left(4x^2+8\right)\\\\3\cdot \:5\left(4x^2+8\right)^4\cdot \:8x\\\\120x\left(4x^2+8\right)^4$

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

3 0

4 years ago

1/1×3 + 1/3×5 + ... + 1/47×49 HELP PLZ

Natalka [10]

Answer:

24/49

Step-by-step explanation:

Let's add the terms and see if there's a pattern

$\dfrac{1}{1\times 3}+\dfrac{1}{3\times 5}=\dfrac{5+1}{1\times 3\times 5}=\dfrac{2}{5}\quad\text{sum of 2 terms}\\\\\dfrac{2}{5}+\dfrac{1}{5\times 7}=\dfrac{14+1}{5\times7}=\dfrac{3}{7}\quad\text{sum of 3 terms}$

Suppose we say the sum of n terms is (n/(2n+1)), the next term in the series will be 1/((2n+1)(2n+3)) and adding that to the presumed sum gives ...

$\dfrac{n}{2n+1}+\dfrac{1}{(2n+1)(2n+3)}=\dfrac{n(2n+3)+1}{(2n+1)(2n+3)}=\dfrac{2n^2+3n+1}{(2n+1)(2n+3)}\\\\=\dfrac{(2n+1)(n+1)}{(2n+1)(2n+3)}=\dfrac{n+1}{2n+3}\text{ matches }\dfrac{(n+1)}{2(n+1)+1}$