Help find the derivative.

Mathematics

1 answer:

Masteriza [31]2 years ago

5 0

Answer:

Step-by-step explanation:

$1.\\f(x)=e^x\\f'(x)=e^x\\2.\\y=x^8-12x^5-3x^2+7x+22\\\frac{dy}{dx}=8x^7-12*5 x^4-3*2x+7+0\\\frac{dy}{dx}=8x^7-60 x^4-6x+7\\3.\\f(x)=e^3\\f'(x)=0$

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The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)

Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

$f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}$

Now, we have to find the quotient of the above expression.

So, here we go:

$Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)$

Now, we will divide the above simplest form with g(x):

$\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2$