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How can this be wrong..I am doing some online course and solving some problems . I am surprised to see this answer . Please guide me in this problem . I simply used liebnitz rule to differentiate under the integral sign.

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    it's not clear to me what you thought the answer was. The answer marked wrong involves no differentiation under the integral sign and the answer marked right uses the Leibniz rule properly – 2017-01-28
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    I have used liebnitz theorem only. I have been using it for a long time now.. is it not correct? – 2017-01-28
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    What is last expression in option $c$? – 2017-01-28
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    This invloves partial derivatives as its a two variable case – 2017-01-28
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    @AbhishekChandra You need to account for both the changes when you vary the endpoints and when you vary the function inside, since all depend on $x$ – 2017-01-28

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The derivative of a function of the form $$F(x) = \int_{a(x)}^{b(x)}f(x,y)dy $$ (where $f$, $a$ and $b$ obey some weak regularity conditions) is given by the Leibniz rule $$F'(x) = b'(x)f(x,b(x))-a'(x)f(x,a(x)) + \int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,y)dy.$$

In your case, $b(x) = x^3$, $a(x)= x^2$ and $f(x,y) = \tan(xy^2)$ so we have $$ F'(x) = 3x^2\tan(x^7) - 2x\tan(x^5) + \int_{x^2}^{x^3}y^2\sec^2(xy^2)dy$$