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Question: If ƒ(x) = \(\int_{1}^{x^{3}}\frac{dt}{1 + t^{4}}\), then ƒ¢¢(x) is equal to -...

If ƒ(x) = ∫1x3dt1+t4\int_{1}^{x^{3}}\frac{dt}{1 + t^{4}}, then ƒ¢¢(x) is equal to -

A

6x(1−5x12)(1+x12)2\frac{6x(1 - 5x^{12})}{(1 + x^{12})^{2}}

B

6x(1+5x12)(1+x12)2\frac{6x(1 + 5x^{12})}{(1 + x^{12})^{2}}

C

– 6x(1–5x12)(1+x12)2\frac{6x(1–5x^{12})}{(1 + x^{12})^{2}}

D

None of these

Answer

6x(1−5x12)(1+x12)2\frac{6x(1 - 5x^{12})}{(1 + x^{12})^{2}}

Explanation

Solution

We have, ƒ(x) = ∫1x3dt1+t4\int_{1}^{x^{3}}\frac{dt}{1 + t^{4}}

Ž ƒ¢(x) = [11+t4]t=x3\left\lbrack \frac{1}{1 + t^{4}} \right\rbrack_{t = x^{3}}. ddx(x3)\frac{d}{dx}(x^{3}) = 11+x12\frac{1}{1 + x^{12}} . 3x2

\ ƒ¢¢(x) = 3 [(1+x12).2x–x2.12x11(1+x12)2]\left\lbrack \frac{(1 + x^{12}).2x–x^{2}.12x^{11}}{(1 + x^{12})^{2}} \right\rbrack

ƒ¢(x) = 6x(1+x12–6x12)(1+x12)2\frac{6x(1 + x^{12}–6x^{12})}{(1 + x^{12})^{2}}

ƒ¢¢(x) = 6x(1–5x12)(1+x12)2\frac{6x(1–5x^{12})}{(1 + x^{12})^{2}}