y=x^2+3x+1/x
2007-06-02 19:45:33 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Show Working Please I am lost?!?
2007-06-02 19:49:28 · update #1
y = x^2 + 3x + 1/x
differentiating both sides wrt x
y' = 2x + 3 - 1/x^2
2007-06-02 20:04:50 · answer #1 · answered by raja 3 · 0⤊ 0⤋
dy = 2 x dx + 3 dx - 1/x² dx
dy = (2x + 3 - 1/x²) dx
y' = dy/dx = (2x + 3 - 1/x²), which is the derivative you are looking for.
Remember the power rule for differentiation:
If y = x^n, then y' = dy/dx = n x^(n-1).
Differentiate each term individually.
(x²)' = 2x dx.
(3x)' = 3 dx.
The last term, 1/x, is actually x^-1. So, using the method above, its derivative is:
(-1) x^(-1-1) = (-1) x^-2 = -1/x².
2007-06-02 20:29:05 · answer #2 · answered by MathBioMajor 7 · 0⤊ 0⤋
y = f(x) = x² + 3x + 1.x^(-1)
dy/dx = f `(x) = 2x + 3 - x^(-2)
dy/dx = f `(x) = 2x + 3 - 1 / x²
2007-06-03 05:06:57 · answer #3 · answered by Como 7 · 0⤊ 0⤋