#### Please solve RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 1 maths textbook solution

$4\pi r+2\pi h$

Hint:

To find the rate of change of total surface area of cylinder, we need to differentiate it with respect to radius

Total surface area of cylinder, $A=2\pi r(r+h)$

Given:

$radius\; (r)\; and \: height\; (h)\; o\! f\; cylinder.$

Solution:

Here we have to find rate of change of the total surface area of cylinder. here radius=$r$, height=$h$
so, total surface area of cylinder.

$A=2\pi r(r+h)$

$\therefore A= 2\pi r^{2}+2\pi rh\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)$

Let’s differentiate equation (i) with respect to radius

$\therefore \frac{d A}{d r}=\frac{d}{d r}\left(2 \pi r^{2}+2 \pi r h\right)$

$\frac{d A}{d r}=\frac{d}{d r}\left(2 \pi r^{2}\right)+\frac{d}{dr}(2\pi rh)$

$\therefore \frac{d A}{d r}=4\pi r+2\pi h$                        $[\because \frac{d (x^{n})}{d x}=nx^{n-1}]$

Which is rate of change of total surface area with respect to radius.

Note:

Here we have took rate with respect to radius because radius was varying.

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