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Valentin [98]
2 years ago
8

First derivative of this function ?​

Mathematics
1 answer:
german2 years ago
7 0

Use the chain rule.

f'(x) = \left[\left(x^2 + \cos^2(x)\right)^3\right]' \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left[x^2 + \cos^2(x)\right]' \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left(\left[x^2\right]' + \left[\cos^2(x)\right]'\right) \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left(2x + 2 \cos(x) \left[\cos(x)\right]'\right) \\\\ ~~~~ = 3 \left(x^2 + \cos^2(x)\right)^2 \left(2x - 2 \cos(x) \sin(x)\right) \\\\ ~~~~ = \boxed{3 \left(x^2 + \cos^2(x)\right)^2 \left(2x - \sin(2x)\right)}

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- 12x - 6x < - 30 - 24
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42 and 165 is 207

Mark me as brainliest if this helps!

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Prove this (sinx-tanx)(cosx-cotx)=(sinx-1)(cosx-1)
Ilya [14]

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<span>=<span>(<span>sinx</span>−<span><span>sinx</span><span>cosx</span></span>)</span><span>(<span>cosx</span>−<span><span>cosx</span><span>sinx</span></span>)</span></span>

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6 0
3 years ago
Read 2 more answers
Mary deposits R400 every three months into a bank account earning interest at 16% per year, compounded quarterly. The approximat
ycow [4]

Answer:

The approximate time that it will take Mary to save R40000 is 29 years

Step-by-step explanation:

A= p(1+r/n)^nt

t = log(A/P) / n[log(1 + r/n)]

A = R40000

n = 4

P = R400

r = 16%

t = log(A/P) / n[log(1 + r/n)]

= log(40,000 / 400) / 4{log(1+0.16/4)}

= log(100) / 4{log(1+0.04)}

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= 2 / 4(0.0170)

= 2 / 0.068

= 29.41

t = 29.41 years

Approximately 29 years

The approximate time that it will take Mary to save R40000 is 29 years

3 0
3 years ago
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