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Question: What is the derivative of \[\pi .{r^2}\]?...

What is the derivative of π.r2\pi .{r^2}?

Explanation

Solution

We need to find the derivative of π.r2\pi .{r^2}. We will be finding the derivative of this expression with respect to rr using the formulas for differentiation. We will be using power rule and property involving constant term, which are as follows:

Formula used:
POWER RULE - ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}
PROPERTY INVOLVING CONSTANT - ddx(c.f(x))=c.(ddx(f(x)))\dfrac{d}{{dx}}\left( {c.f\left( x \right)} \right) = c.\left( {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)} \right), where cc is a constant term.

Complete step by step answer:
We need to find the derivative of π.r2\pi .{r^2} with respect to rr.
Let y=π.r2−−−−−−−(1)y = \pi .{r^2} - - - - - - - (1)
So, we need to find dydr\dfrac{{dy}}{{dr}}
Differentiating (1) with respect to rr, we get
⇒dydr=ddr(π.r2)\Rightarrow \dfrac{{dy}}{{dr}} = \dfrac{d}{{dr}}\left( {\pi .{r^2}} \right)

Here we see that π\pi is a constant term. Hence using the Property
ddx(c.f(x))=c.(ddx(f(x)))\dfrac{d}{{dx}}\left( {c.f\left( x \right)} \right) = c.\left( {\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)} \right), where cc is a constant term, we get
⇒dydr=π.(ddrr2)\Rightarrow \dfrac{{dy}}{{dr}} = \pi .\left( {\dfrac{d}{{dr}}{r^2}} \right)

Now using the Property ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}, we get
⇒dydr=π.(2r2−1)\Rightarrow \dfrac{{dy}}{{dr}} = \pi .\left( {2{r^{2 - 1}}} \right)
⇒dydr=π.(2r1)\Rightarrow \dfrac{{dy}}{{dr}} = \pi .\left( {2{r^1}} \right)
⇒dydr=π.(2r)\Rightarrow \dfrac{{dy}}{{dr}} = \pi .\left( {2r} \right)
As multiplication is commutative we can write the above expression as
∴dydr=2πr\therefore \dfrac{{dy}}{{dr}} = 2\pi r

Hence derivative of π.r2\pi .{r^2} with respect to rr is 2πr2\pi r.

Note: We can also solve the given problem using First Principle of Differentiation.According to First Principle of differentiation, the derivative of a function f(x)f\left( x \right) can be evaluated by calculating the limit f′(r)=lim⁡h→0f(r+h)−f(r)hf'\left( r \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {r + h} \right) - f\left( r \right)}}{h}, where f′(r)f'\left( r \right) is the first derivative of the function f(r)f\left( r \right) with respect to rr.