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JEE Main & Advanced Sample Paper JEE Main Sample Paper-15

  • question_answer
    Let \[f'(x)=\sin \,{{x}^{2}}\] and \[y=f({{x}^{2}}+1)\] then \[\frac{dy}{dx}\]at \[x=2\] is

    A)  0                                            

    B)  1

    C)  4 sin 5                  

    D)  4 sin 25

    Correct Answer: D

    Solution :

    \[f'(x)=\sin {{x}^{2}}\]\[y=f({{x}^{2}}+1)\] \[\frac{dy}{dx}=f'({{x}^{2}}+1)\cdot 2x\] \[{{\left. \frac{dy}{dx} \right|}_{x=2}}=4f'(5)=4\sin 25.\]


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