MCQ
${d \over {dx}}{({x^2} + \cos x)^4} = $
- A$4({x^2} + \cos x)(2x - \sin x)$
- B$4{({x^2} - \cos x)^3}(2x - \sin x)$
- ✓$4{({x^2} + \cos x)^3}(2x - \sin x)$
- D$4{({x^2} + \cos x)^3}(2x + \sin x)$
✓
Answer
Correct option: C.
$4{({x^2} + \cos x)^3}(2x - \sin x)$
c
(c) $\frac{d}{{dx}}{[{x^2} + \cos x]^4} = 4{[{x^2} + \cos x]^3}[2x - \sin x]$.
(c) $\frac{d}{{dx}}{[{x^2} + \cos x]^4} = 4{[{x^2} + \cos x]^3}[2x - \sin x]$.
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