Differentiate (x2−5x+8)(x3+7x+9)in three ways mentioned below.
(a) By using product rule.
(b) By expanding the product to obtain a single polynomial.
(c) By logarithmic differentiation.
Do they all given the same answer?
Ley y=(x2−5x+8)(x3+7x+9) .,........(i)
(a) Using product rule,
dydx=(x2−5x+8)ddx(x3+7x+9)+(x3+7x+9)ddx(x2−5x+8) =(x2−5x+8)(3x2+7)+(x3+7x+9)(2x−5) =3x4−15x3+24x2+7x2−35x+56+2x4+14x2+18x−5x3−35x−45........(ii) =5x4−20x3+45x2−52x+11
(b) Writing y as a single polynomial,
y=(x2−5x+8)(x3+7x+9)=x5−5x4+15x3−26x2+11x+72Differentiating both sides w.r.t. x,we get∴ dydx=ddx(x5−5x4+15x3−26x2+11x+72) =5x4−20x3+45x2−52x+11 ......(iii)
(c) Taking logarithms on both sides of Eq. (i), we get
log y=log{(x2−5x+8)(x3+7x+9)}⇒ log y=log{(x2−5x+8)+log(x3+7x+9)}Differentiating both sides w.r.t. x,we get1ydydx=1x2−5x+8ddx(x2−5x+8)+1x3+7x+9ddx(x3+7x+9) (Using chain rule)1ydydx=2x−5x2+5x+8+3x2+7x3+7x+9 =(x2−5x+8)(x3+7x+9)(2x−5x2−5x+8+3x2+7x3+7x+9)=(2x−5)(x3+7x+9)+(3x2+7)(x2−5x+8)=2x4+14x2+18x−5x3−35x−45+3x4−15x3+24x2+7x2−35x+56=5x4−20x3+45x2−52x+11 .........(iv)
From Eqs. (ii), (iii) and (iv), we find that result is same in every case.