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Mathematics

2 answers:

Solnce55 [7]3 years ago

8 0

Answer:

The common facor is x-3.

Step-by-step explanation:

The first expression can be factored with a difference of squares, x+3 and x-3. You can check that by multiplying them together to get x²-9. Knowing that you need a common factor, one that works for both expressions, you can immediately eliminate A and B. The second expression, x²-5x+6, has to share one of these two factors. To factor this, you find the numbers that multiply to the c term and add to the b term. (Ax²+bx+c) so b is -5 and c is 6. All you have to do is find out which two numbers add to -5 and multiply to 6. You start with factors of 6, which are +-6 and +-1, and +-3 and +-2. Say you try 6 and 1 first. The b term is negative, so these two added have to add to a negative number, which means positive 6 and positive 1 won´t work. The only way to add to -5 is to add 1 to -6 or -6 to 1, but multiplying those together gives you -6, and you need positive 6 in the c term. That leaves you with -2 and -3. they add to -5, and multiply to -6, which means that your common term is x-3. Test this by expanding your result, (x-2)(x-3), if you get the original expression theyre both factors.

EleoNora [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Both equations use x squared

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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

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