Differentiate (x2 -- 5x + 8) (x' + 7x + 9) in three ways mentioned belo
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(ii) by logarithmic differentiation.
Do they all give the same answer?
Answers
Answer:
ANSWER
Let y=(x
2
−5x+8)(x
3
+7x+9)
(i) Let x
2
−5x+8=u and x
3
+7x+9=v
∴y=uv
⇒
dx
dy
=
dx
du
.v+u.
dx
dv
By using product rule,
⇒
dx
dy
=
dx
d
(x
2
−5x+8).(x
3
+7x+9)+(x
2
−5x+8).
dx
d
(x
3
+7x+9)
⇒
dx
dy
=(2x−5)(x
3
+7x+9)+(x
2
−5x+8)(3x
2
+7)
⇒
dx
dy
=2x(x
3
+7x+9)−5(x
3
+7x+9)+x
2
(3x
2
+7)−5x(3x
2
+7)+8(3x
2
+7)
⇒
dx
dy
=(2x
4
+14x
2
+18x)−5x
3
−35x−45+(3x
4
+7x
2
)−15x
3
−35x+24x
2
+56
∴
dx
dy
=5x
4
−20x
3
+45x
2
−52x+11
(ii) y=(x
2
−5x+8)(x
3
+7x+9)
=x
2
(x
3
+7x+9)−5x(x
3
+7x+9)+8(x
3
+7x+9)
=x
5
+7x
3
+9x
2
−5x
4
−35x
2
−45x+8x
3
+56x+72
=x
5
−5x
4
+15x
3
−26x
2
+11x+72
∴
dx
dy
=
dx
d
(x
5
−5x
4
+15x
3
−26x
2
+11x+72)
=
dx
d
(x
5
)−5
dx
d
(x
4
)+15
dx
d
(x
3
)−26
dx
d
(x
2
)+11
dx
d
(x)+
dx
d
(72)
=5x
4
−5×4x
3
+15×3x
2
−26×2x+11×1+0
=5x
4
−20x
3
+45x
2
−52x+11
(iii) y=(x
2
−5x+8)(x
3
+7x+9)
Taking logarithm on both the sides, we obtain
logy=log(x
2
−5x+8)+log(x
3
+7x+9)
Differentiating both sides with respect to x, we obtain
y
1
dx
dy
=
dx
d
log(x
2
−5x+8)+
dx
d
log(x
3
+7x+9)
⇒
y
1
dx
dy
=
x
2
−5x+8
1
.
dx
d
(x
2
−5x+8)+
x
3
+7x+9
1
.
dx
d
(x
3
+7x+9)
⇒
dx
dy
=y[
x
2
−5x+8
1
×(2x−5)+
x
3
+7x+9
1
×(3x
2
+7)]
⇒
dx
dy
=(x
2
−5x+8)(x
3
+7x+9)[
x
2
−5x+8
2x−5
+
x
3
+7x+9
3x
2
+7
]
⇒
dx
dy
=(x
2
−5x+8)(x
3
+7x+9)[
(x
2
−5x+8)(x
3
+7x+9)
(2x−5)(x
3
+7x+9)+(3x
2
+7)(x
2
−5x+8)
]
⇒
dx
dy
=2x(x
3
+7x+9)−5(x
3
+7x+9)+3x
2
(x
2
−5x+8)+7(x
2
−5x+8)
⇒
dx
dy
=(2x
4
+14x
2
+18x)−5x
3
−35x−45+(3x
4
−15x
3
+24x
2
)+(7x
2
−35x+56)
⇒
dx
dy
=5x
4
−20x
3
+45x
2
−52x+11
From the above three observations, it can be concluded that all the
dx
dy
results of
Answer:
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