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if `y=x^2sinx+(3x)/(tanx)`, then `(dy)/(dx)` will beA. `2xsinx+x^2cosx+(3tanx-3xsec^2x)/(tan^2x)`B. `2xsinx+(3xsec^2x-3tanx)/(tan^2x)`C. `(x^2cosx+2xsinx-(3tanx-3xsec^2x))/(tan^2x)`D. `(x^2cosx-2xsinx-(3tanx-3xsec^2x))/(tan^2)` |
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Answer» Correct Answer - A `y=x^2sinx+(3x)/(tanx)` `(dy)/(dx)=x^2(d)/(dx)sinx+sinx(d)/(dx)(x^2)` `+((tan)(d)/(dx)(3x)-3x(d)/(dx)(tanx))/(tan^2x)` `(dy)/(dx)=x^2cosx+2xsinx+(3tanx-3xsec^2x)/(tan^2x)` |
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