Basic calculus-1

Calculus Level 2

y = x 3 − 27 x 2 + 3 x + 9 \large y=\frac{x^3-27}{x^2+3x+9}

Find d y d x \dfrac {dy}{dx} .

0 2 4 5 1 3

Mohammad Khaza by Mohammad Khaza , 20, Bangladesh

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1 solution

Mohammad Khaza
Oct 4, 2017

given that, y = x 3 − 27 x 2 + 3 X + 9 y=\frac{x^3-27}{x^2+3X+9}

shaping that, y = x 3 − 3 3 x 2 + 3 x + 9 y=\frac{x^3-3^3}{x^2+3x+9}

or, y = ( x − 3 ) ( x 2 + 3 x + 9 ) ( x + 3 x + 9 ) y=\frac{(x-3)(x^2+3x+9)}{(x^+3x+9)}

or, y = x − 3 y=x-3

now, d ( x − 3 ) d x = 1 − 0 = 1 \frac{d(x-3)}{dx}=1-0=1

5 Helpful 0 Interesting 0 Brilliant 0 Confused

It should be checked that the denominator has no zero solutions. This is guaranteed if x is real, because D = 3 3 − 4 ⋅ 1 ⋅ 9 < 0 D=3^3-4 \cdot 1 \cdot 9<0 , but in the complex domain there are singularities at x = − 3 2 ± 3 2 3 i x=-\frac{3}{2} \pm \frac{3}{2}\sqrt{3}i for which the function value and its derivative are undefined.

K T - 2 years ago

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