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2 years ago

Can someone tell me to stop slacking off school please LOL

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

5 0

Stop slacking off and be a bad a

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Find the derivative of f(x)= (e^ax)*(cos(bx)) using chain rule

Vikentia [17]

$f(x) = e^{ax}\cos(bx)$

then by the product rule,

$f'(x) = \left(e^{ax}\right)' \cos(bx) + e^{ax}\left(\cos(bx)\right)'$

and by the chain rule,

$f'(x) = e^{ax}(ax)'\cos(bx) - e^{ax}\sin(bx)(bx)'$

which leaves us with

$f'(x) = \boxed{ae^{ax}\cos(bx) - be^{ax}\sin(bx)}$

Alternatively, if you exclusively want to use the chain rule, you can carry out logarithmic differentiation:

$\ln(f(x)) = \ln(e^{ax}\cos(bx)} = \ln(e^{ax})+\ln(\cos(bx)) = ax + \ln(\cos(bx))$

By the chain rule, differentiating both sides with respect to x gives

$\dfrac{f'(x)}{f(x)} = a + \dfrac{(\cos(bx))'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a - \dfrac{\sin(bx)(bx)'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a-b\tan(bx)$

Solve for f'(x) yields

$f'(x) = e^{ax}\cos(bx) \left(a-b\tan(bx)\right) \\\\ f'(x) = e^{ax}\left(a\cos(bx)-b\sin(bx))$

just as before.

4 0

2 years ago

What is 3/8 + 1/8 - 2/7 + 1/4

MatroZZZ [7]

0.46428571428 you have to know how to add and subtracts fractions

6 0

3 years ago

Read 2 more answers

FIRST CORRECT ANSWER GETS BRAINLIEST!!!

Luba_88 [7]

If two secants are drawn from a point lying outside a circle, then the product of one secant on its outer part is equal to the product of the other secant on its outer part.

DE*CE = AE*BE

DE = CD + CE = 1 + x + 4 = x + 5
AE = AB + BE = 11 + x + 1 = x + 12

(x+5)(x+4)=(x+12)(x+1)
x² + 4x + 5x + 20 = x² + x + 12x + 12
9x + 20 = 13x + 12
20 - 12 = 13x - 9x
8 = 4x
x = 8/4
x = 2

4 0

2 years ago

2 times as many as 7 is

Bogdan [553]

Answer: 14

Step-by-step explanation:

Basically 7x2

=14

5 0

3 years ago

Divide £144 in the ratio 7:5

prohojiy [21]

84$ to 60$ is the answer

7 0

3 years ago