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Question
Differentiate the following w.r.t.x:
(x3 – 2x – 1)5
Solution 1
Let y = (x3 – 2x – 1)5
Differentiating w.r.t.x, we get
`"dy"/"dx" = "d"/"dx"` (x3 – 2x – 1)5
= 5(x3 – 2x – 1)4 × `"d"/"dx"` (x3 – 2x – 1)
= 5(x3 – 2x – 1)4 × (3x2 – 2 × 1 – 0)
= 5(3x2 – 2)(x3 – 2x – 1)4.
Solution 2
Let y = (x3 – 2x – 1)5
Put u = x3 – 2x – 1.
Then y = u5
∴ `"dy"/"du" = "d"/"du"` (u5) = 5u4
= 5(x3 – 2x – 1)4
and
`"du"/"dx" = "d"/"dx"`(x3 – 2x – 1)
= 3x2 – 2 × 1 – 0
= 3x2 – 2
∴ `"dy"/"dx" = "dy"/"du" xx "du"/"dx"`
= 5(x3 – 2x – 1)4 (3x2 – 2)
= 5(3x2 – 2)(x3 – 2x – 1)4.
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