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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 51 of 72
Marks: +1, -0
ax+bpx2+qx+r\frac{ax+b}{px^{2}+qx+r}
Solution:  
Let f (x) = ax+bpx2+qx+r\frac{ax+b}{px^{2}+qx+r} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} (f (x)) =
(px2+qx+r)(ax+b)′−(ax+b)(px2+qx+r)′(px2+qx+r)2\frac{(px^{2}+qx+r)(ax+b)'-(ax+b)(px^{2}+qx+r)'}{(px^{2}+qx+r)^{2}}
=
(px2+qx+r)(a+0)−(ax+b)(2px+q+0)(px2+qx+r)2\frac{(px^{2}+qx+r)(a+0)-(ax+b)(2px+q+0)}{(px^{2}+qx+r)^{2}}
=
px2a+qxaa+ra−2pax2−axq−2pxb−bq(px2+qx+r)2\frac{px^{2}a+qxaa+ra-2pax^{2}-axq-2pxb-bq}{(px^{2}+qx+r)^{2}}
=
−pax2+ra−2pxb−bq(px2+qx+r)2\frac{-pax^{2}+ra-2pxb-bq}{(px^{2}+qx+r)^{2}}
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