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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 52 of 72
Marks: +1, -0
px2+qx+rax+b\frac{px^2+qx+r}{ax+b}
Solution:  
Let f (x) = px2+qx+rax+b\frac{px^2+qx+r}{ax+b}
⇒ f (x) = (px2+qx+r)(ax+b)−1(px^2+qx+r)(ax+b)^{-1} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} (f (x)) = (2px + q) (ax+b)−1(ax+b)^{-1} + (px2+qx+r)[(−1)(ax+b)−2a](px^2+qx+r)[(-1)(ax+b)^{-2}a]
= 2px+qax+b\frac{2px+q}{ax+b} - a(px2+qx+r)(ax+b)2\frac{a(px^2+qx+r)}{(ax+b)^2}
=
(2px+q)(ax+b)−apx2−aqx−ar(ax+b)2\frac{(2px+q)(ax+b)-apx^2-aqx-ar}{(ax+b)^2}
=
2px2a+2pxb+aqx+bq−apx2−aqx−ar(ax+b)2\frac{2px^2 a + 2pxb + aqx + bq - apx^2 - aqx - ar}{(ax+b)^2}
=
apx2+2pxb+bq−ar(ax+b)2\frac{apx^2 + 2pxb + bq - ar}{(ax+b)^2}
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