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Please solve RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 5 maths textbook solution

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\frac{2}{3}\pi rh


Here we have to differentiate volume of cone (V) with respect to radius of base (r)

V\! olume\; o\! f\; cone, V=\frac{1}{3}\pi r^{2}h

Where h is height of cone.


Volume of cone.


Here we have,

r = Radius of base of cone

h = Height of cone

V\! olume\; o\! f\; cone, V=\frac{1}{3}\pi r^{2}h

Let’s differentiate Vwith respect to r,

\therefore \frac{dV}{dr}=\frac{2}{3}\pi rh                                                \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]


Here height (h) is constant so we don’t have to differentiate it.

Posted by

Gurleen Kaur

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