We have,
f'(x) = 4x³+32 = 0 gives x³ = -8
⇒ x = −2
and f′′(x) = 12 x² 
As f''(−2)>0, the function has local minimum at x = −2. The local minimum value is f (−2) = −48 
Therefore, the extreme point is (−2, −48) .

Let x be the length of the garden and y be the breadth of the garden.

Given 2 (x +y) = 40
[ஃ length of the wire = 40 Perimeter = 40 m]
⇒ x + y = 20
⇒ y = 20 - x ...(1)
Let f(x) = xy = x (20 - x²) = 20x- xl²
f'(x) = 20 - 2x
f'(x) = 0
⇒ 20-2x = 0
⇒ 20 = 2x
x = 10
ஃ The critical number is 10
f"(x) = -2
Now f"(10) = -2 < 0
ஃ f(x) is minimum at x = 10
When x = 10, y = 20 -10 = 10
ஃ Area = f(x) = xy
= 10(10) = 100m²
ஃ Largest possible area of the garden = 100 m²

Given x + y = 10
⇒ y = 10 -x ...(1)
Let f(x) = x²+y
= x²+ (10 -x)²
= x + 100 +x²- 20x
f(x) = 2x² - 20x+ 100
f'(x) = 4x - 20
f'(x) = 0
4x-20 = 0
4x = 20
⇒ x = 5
∴The critical number is 5
f"(x) = 4
ஃ f"(5) = 4 > 0
ஃf(x) is minimum when x = 5
When x = 5, y = 10- 5 = 5
[From (1)]
ஃ Smallest possible value of x²+y²
= 5² + 5² = 25 + 25 = 50

Given f(x) = 3x² - 4x³ ; [-1, 2]
f'(x) = 12x³ - 12x²
f'(x) = 0
⇒12x³- 12x² = 0
⇒ 12x²(x-1) = 0
⇒ x = 0 or x = 1
Evaluatingf(x) at the end points x = -1, x = 2 and at the critical number x = 0, x = 1 we get
f(-1) = 3(-1)4-4 (-1)³
= 3 + 4 = 7
f(2) = 3(2)⁴ - 4(2³)
= 48 - 32 =16
f(0) = 0
f(1) = 3(1)⁴ - 4(1)³
= 3 - 4 = -1
From these values, the absolute maximum is 16 at x = 2 and the absolute minimum is -1 which occurs at x = 1.

The average rate of change in an interval [a, b] is \(\frac { f(b)-f(a) }{ b-a } \) whereas, the instantaneous rate of change at a point x is f′(x) for the given function. They are respectively, b + a and 2x.

We have,
\(f'(x)=2x>0, \forall x\in(2,7)\) and
\(f'(x)=2x>0, \forall x\in(-2,0)\)
and hence the proof is completed.

This is an indeterminate of the form \((\frac{\infty}{\infty})\)
To evaluate this limit, we apply l’Hôpital Rule m times
\(\underset{x\rightarrow \infty}{lim}\frac{e^{x}}{x^{m}}=\underset{x\rightarrow \infty}{lim}\frac{e^{x}}{m!} = \infty\) 

This is an indeterminate of the form (0×∞). To evaluate this limit, we first simplify and bring it to the form \((\frac{\infty}{\infty})\) and apply L’Hôpital Rule.
\(\underset{x\rightarrow 0^{+}}{lim}xlog x= \underset{x\rightarrow 0^{+}}{(\frac{log x}{\frac{1}{x}})}\)
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{\frac{1}{x}}{-\frac{1}{x^{2}}})=\underset{x\rightarrow 0^{+}}{lim}(-x)=0.\)

This is an indeterminate of the form \((\frac{\infty}{\infty})\).
To evaluate this limit, we apply l ’Hôpital Rule.
\(\underset{x\rightarrow \infty }{lim}(\frac{x^{2}+17x+29}{x^{4}})=\underset{x\rightarrow \infty}{lim}(\frac{2x+17}{4x^{3}})\)
= \(\underset{x\rightarrow \infty}{lim}(\frac{2}{12x^{2}})=0\)

As this is an indeterminate form \((\frac{0}{0})\) using the l’Hôpital’s Rule
\(\underset{\theta \rightarrow 0}{lim} (\frac{1-cosm\theta}{1-cosn\theta})\)=\(\underset{\theta \rightarrow 0}{lim} (\frac{m-sinm\theta}{n-sinn\theta})\)
\(\underset{\theta\times 0}{lim} \frac{m}{n} (\frac{\frac{sin \ m\theta}{\theta}}{\frac{sin \ n\theta}{\theta}}) =\frac{m^{2}}{n^{2}}\)
Therefore, m² = n²
That is \(m=\pm n\).

Let f(x) = log x
f^l(x) = \(\frac{1}{x}\) = x^-1
f^lI(x) = -1x^-2
f^III(x) = 2x^-3
f^IV(x) = -6 x^-4
⇒ f(1) = log 1 = 0
f^l(1) = \(\frac11\) = 1
f^ll(1) = -1
f^lll(1) = 2
f^IV(1) = -6
Taylor series for f(x) at x = 1 is
\(f(x)=f(1)+\frac { { f }^{ 1 }(1) }{ 1! } (x-1)+\frac { { f }^{ II }(1) }{ 2! } ({ x-1) }^{ 2 }+\frac { { f }^{ III }(1) }{ 3! } { (x-1) }^{ 3 }+\).....
log x = \(\frac { 1 }{ 1! } (x-1)+\frac { -1 }{ 2! } { (x-1) }^{ 2 }+\frac { 2 }{ 3! } ({ x-1) }^{ 3 }-\frac { 6 }{ 4! } ({ x-1) }^{ 4 }+..\)
log x = \((x-1)-\frac { 1 }{ 2 } { (x-1) }^{ 2 }+\frac { 1 }{ 3 } { (x-1) }^{ 3 }-\frac { 1 }{ 4 } ({ x-1) }^{ 4 }\)+ ....

Let the two positive numbers be x and y.
Given xy = 20
\(\Rightarrow y=\frac { 20 }{ x } \) 
Let f(x) = x+y
\(f(x)=x+\frac { 20 }{ x } \) 
\(f'(x)=1-\frac { 20 }{ { x }^{ 2 } } \)
f'(x) = 0
\(\Rightarrow 1-\frac { 20 }{ { x }^{ 2 } } =0\) 
\(\Rightarrow 1=\frac { 20 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=20\) 
\(\Rightarrow x=\pm \sqrt { 20 } \) 
\(\Rightarrow x=\pm 2\sqrt { 5 } \)
∴ The critical number are \(2\sqrt { 5 } -2\sqrt { 5 } \) 
\(f''(x)=\frac { 40 }{ { x }^{ 3 } } \) 
When \(x=2\sqrt { 5 } \)
\(f''\left( x \right) =\frac { 40 }{ \left( 2\sqrt { 5 } \right) ^{ 3 } } >0\) 
ஃ f(x) is minimum when \(x=2\sqrt { 5 } \) 
When \(x=2\sqrt { 5 } ,y=\frac { 20 }{ 2\sqrt { 5 } } \) 
= \(\frac { 10 }{ \sqrt { 5 } } \times \frac { \sqrt { 5 } }{ \sqrt { 5 } } =\frac { 10\sqrt { 5 } }{ 5 } =2\sqrt { 5 } \) 
Hence the required positive numbers are \(2\sqrt { 5 } \)
Minimum sum = \(2\sqrt { 5 } \) + \(2\sqrt { 5 } \)  = \(4\sqrt { 5 } \)

f'(x) = -2 sin x + 2 cos 2x
f'(x) = 0
\(\Rightarrow\) 2 sin x + 2 cos 2x = 0
\(\Rightarrow\) 2 sin x + 2(1 - 2 sin²x) = 0
\(\Rightarrow\) 4 sin²x + 2 sin x - 2 = 0
\(\Rightarrow\) 4 sin² x + 2 sin x - 2 = 0
\(\Rightarrow\) 2 sin² x + sin x-1 = 0
\(\Rightarrow\) (sin x+1) (2sin x-1) = 0
\(\Rightarrow\) \(sinx=-1\ or\ sinx=\frac { 1 }{ 2 } \) 
\(\Rightarrow\) \(sinx=-sin\frac { \pi }{ 2 } \) or 
\(sinx=sin\frac { \pi }{ 6 } \) 
 \(\Rightarrow sinx=sin\left( -\frac { \pi }{ 2 } \right) \) 
\(sinx=sin\frac { \pi }{ 6 } \)  
\(\Rightarrow x=-\frac { \pi }{ 2 } or \ x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=\frac { \pi }{ 6 } \) 
\(\\ \\ \\ \\ \\ \\ \left[ \because x=-\frac { \pi }{ 2 }∉ \left[ 0.\frac { \pi }{ 2 } \right] \right] \)
\(\therefore\) The critical number is \(x=\frac { \pi }{ 6 } \) 
Evaluating f(x) at the end points x = 0, \(x=\frac { \pi }{ 2 } \) and at the critical number \(x=\frac { \pi }{ 6 } \) we get.
f(0) = 2 cos0 + sin0 = 2
\(f(\frac { \pi }{ 2 } )=2cos\frac { \pi }{ 2 } +sin\pi =0\) 
\(f\left( \frac { \pi }{ 6 } \right) =2cos\frac { \pi }{ 2 } +sin\frac { \pi }{ 3 } \) 
\(2\left( \frac { \sqrt { 3 } }{ 2 } \right) +\frac { \sqrt { 3 } }{ 2 } =\frac { 3\sqrt { 3 } }{ 2 } \) 
From these values, the absolute maximum is \(\frac { 3\sqrt { 3 } }{ 2 } \) which occurs at \(x=\frac { \pi }{ 6 } \) and the absolute minimum is 0 which occurs at \(x=\frac { \pi }{ 2 } \)

f (1) = 0 and f(2) = -2. Clearly f (x) is defined and differentiable in 1< x <2.
Therefore,  by the Mean Value Theorem, there exists a c∈(1, 2) such that by
\(f'(c)=\frac{f(2)-f(1)}{2-1}=1-2c\)
That is  \(1-2c= -2 \Rightarrow c= \frac{3}{2}\)

\(f'\left( x \right) =6\times \frac { 4 }{ 3 } { x }^{ \frac { 4 }{ 3 } -1 }-3\times \frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } -1 }\) 
= \({ 8x }^{ \frac { 1 }{ 3 } }-x^{ \frac { -2 }{ 3 } }\) 
f'(x) = 0
\(\Rightarrow{ 8 }x^{ \frac { 1 }{ 3 } }-\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow \frac { 8x-1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow x=\frac { 1 }{ 8 } \)
Thus, the critical number is \(x=\frac { 1 }{ 8 } \) 
Evaluating f(x) at the end points = -1
x = 1 and at the critical number x = \(\frac { 1 }{ 8 } \) 
we get
\(f(-1)=6(-1)^{ \frac { 4 }{ 3 } }-3\left( -1 \right) ^{ \frac { 1 }{ 3 } }\) 
= 6( 1) - 3 (-1) = 6 + 3 = 9
\(f(1)=6(1)^{ \frac { 4 }{ 3 } }-3(1)^{ \frac { 1 }{ 3 } }=6-3=3\)
\(f\left( \frac { 1 }{ 8 } \right) =6\left( \frac { 1 }{ 8 } \right) ^{ \frac { 4 }{ 3 } }-3\left( \frac { 1 }{ 8 } \right) ^{ \frac { 1 }{ 3 } }\) 
=\(6\left( { 2 }^{ -3 } \right) ^{ \frac { 4 }{ 3 } }-3\left( 2^{ -3 } \right) ^{ \frac { 1 }{ 3 } }\) 
= \(\frac { 6 }{ 16 } -\frac { 3 }{ 2 } =\frac { 3 }{ 8 } -\frac { 3 }{ 2 } \) 
= \(\frac { 3-12 }{ 8 } =\frac { -9 }{ 8 } \)
From these values, the absolute maximum is 9 which occurs at x = -1 and the absolute minimum is \(-\frac { 9 }{ 8 } \) which occurs at x = \(\frac { 1 }{ 8 } \)

Let P(x) = \(a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\). Let  \(\alpha<\beta \) be two real zeros of P(x). Therefore, \(P(\alpha)=P(\beta)=0.\) Since P(x) is continuous in \([\alpha, \beta]\) and differentiable in \((\alpha, \beta)\) by an application of Rolle’s theorem there exists \(\gamma \in (\alpha,\beta)\) such that \(P'(\gamma)=0\). Since,
\(P'(x)=na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+...+a_{1}\) which completes the proof.

f(x) = x² -12x + 10; [1, 2]
Given f(x) = x² -12x + 10 ; [1, 2]
f'(x) = 2x - 12
f'(x) = 0
\(\Rightarrow\) 2x-12 = 0
\(\Rightarrow\) 2x = 12
\(\Rightarrow\) x = 6
\(\therefore\) The critical number is 6
Evaluating f (x) at the end points x = 1,
x = 2 and at the critical number x = 6 we get
f(1) = 1²-12(1)+10 = -1
f(2) = 2²-12(2)+10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10

The curve y = sin x intersects the positive x -axis. When y = 0 which gives, x = 
\( x=n\pi , n=1,2,3,...\)
Now, \(\frac{dy}{dx}=cos x\). The slpoe \(x=n\pi\)  are \(cos(n\pi)=(-1)^{n}\).   
Hence, the required angle of intersection is m₂ = 0
\(tan \theta = \frac{(-1)^n - 0}{1+((-1)^n(0)} = 1 ∀ n\)

The slope of the line y = x is 1. The tangent to the given curve will be parallel to the line, if the slope of the tangent to the curve at a point is also 1. Hence,
\(\frac{dy}{dx}=3x^{2}-6x+1=1\)
which gives \( 3x^{2}-6x=0\)
Hence, x = 0 and x = 2.
Therefore, at (0, –2) and (2, –4) the tangent is parallel to the line y = x.

Given m (x) = \(\sqrt { 3 } x\) = \(\sqrt { 3 } .{ x }^{ \frac { 1 }{ 2 } }\)
Differentiating with respect to 'x' we get,

when x = 3, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 3 } } =\frac { 1 }{ 2 } \) Kg/m
when x = 27, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 27 } } \)
\(=\frac{\sqrt{\not 3}}{2(3) \sqrt{\not 3}}=\frac{1}{6} \mathrm{Kg} / \mathrm{m}\)

Given s = 2t² + 3t
s(3) = 2 \(\times\) 3² + 3 (3)
= 2\(\times\)9+9
= 27 m ....(1)
s(6) = 2\(\times\) 6² + 3 (6)
= 72 + 18 = 90m ... (2)
Average velocity = \(\frac { s(6)-s(3) }{ 6-3 } \)
= \(\frac { 90-27 }{ 3 } \) = 21 m/s
(ii) Instantaneous Velocity V(t) = \(\frac { ds }{ dt } \)
Instantaneous Velocity at t = 3
= V(3) = 15 m/sec  
Instantaneous Velocity at t = 6
= V(6) = 27m/sec

We have,
\(\underset { x\rightarrow { 0 }^{ - } }{ lim } =-\infty \ and\ \underset { x\rightarrow { 0 }^{ x } }{ lim } =\frac { 1 }{ x } =\infty \). Hence, the required vertical asymptote is x = 0 or the y -axis.
As the curve is symmetric with respect to both the axes, y = 0 or the x -axis is also an asymptote.
Hence this (rectangular hyperbola) curve has both the vertical and horizontal asymptotes.

Since f(x) = x² - 2x - 3 , \(f'(x)=2x-2>0 \forall x\in (2, \infty)\). Hence f (x) is strictly increasing in \((2, \infty)\)

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as,
\(\underset{x\rightarrow 0^{+}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{+}}{lim}(\frac{cos \ x}{2x})=\infty\)
\(\underset{x\rightarrow 0^{-}}{lim}(\frac{sin \ x}{x^{2}})=\underset{x\rightarrow 0^{-}}{lim}(\frac{cos \ x}{2x})=\infty\)
As the left limit and the right limit are not the same we conclude that the limit does not exist.
Remark
One may be tempted to use the l’Hôpital’s rule once again in \(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\) to conclude
\(\underset{x\rightarrow 0^{+}}{lim} (\frac{cos \ x}{2x})\)\(\underset{x\rightarrow 0^{+}}{lim} (\frac{-sin \ x}{2})\)=0
which is not true because it was not an indeterminate form.

If we directly substitute x = 0 we get an indeterminate form \(\frac{0}{0}\) and hence we apply the l’Hôpital’s rule to evaluate the limit as
\(\underset{x\rightarrow 0}{lim}(\frac{sin \ mx}{x})\)=\(\underset{x\rightarrow 0}{lim}(\frac{m\times cos \ mx}{1})\)
= m
The next example tells that the limit does not exist.

Given
\(f(x)=\frac { { x }^{ 2 }+6x-4 }{ 3x-6 } \)
\(\underset { x\rightarrow { 2 }^{ + } }{ lim } \frac { { x }^{ 2 }+6x-4 }{ 3x-6 } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { (2+h)^{ 2 }+6(2+h)-4 }{ 3(2+h)-6 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (2+h)^{ 2 }+6(2+h)-4 }{ 6+3h-6 } \)
= -∞
ஃ x = 2 is the vertical asymptote,
Also

\(\therefore y=\frac { 1 }{ 3 } x+\frac { 8 }{ 3 } \) is the slanting asymptote.

If we put directly x = a we observe that the given function is in an indeterminate form \(\frac00\). 
As the numerator and the denominator functions are polynomials they both are differentiable.
Hence by an application of the l’Hôpital Rule we get,
\(\underset{x\rightarrow a}{lim}(\frac{x^{n}-a^{n}}{x-a})\) = \(\underset{x\rightarrow a}{lim}(\frac{n\times x^{n-1}}{1})\)
= \(n \times a^{n-1} \).

If we put directly x = 1 we observe that the given function is in an indeterminate form \(\frac{0}{0}\). As the numerator and the denominator functions are polynomials of degree 2 they both are differentiable.
Hence, by an application of the l’Hôpital Rule, we get
\(\underset{x\rightarrow 1}{lim}(\frac{x^{2}-3x+2}{x^{2}-4x+3})=\underset{x\rightarrow 1}{lime}(\frac{2x-3}{2x-4})\)
= \(\frac{1}{2}\)
Note that this limit may also be evaluated through the factorization of the numerator and denominator as \(\frac{x^{2}-3x+2}{x^{2}-4x+3}=\frac{(x-1)(x-2)}{(x-1)(x-3)}\) 

Given \(f(x)=\frac { { x }^{ 2 }-6x-1 }{ x+3 } \)
\(\underset { x\rightarrow -3^{ + } }{ lim } \frac { { x }^{ 2 }-6x-1 }{ x+3 } =\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( -3+h \right) ^{ 2 }-6(-3+h)-1 }{ -3+h+3 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { 9+{ h }^{ 2 }-6h+18-6h-1 }{ h } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { { h }^{ 2 }-12h+26 }{ h } =\infty \)
and \(\underset { x\rightarrow 3^{ - } }{ lim } \frac { { x }^{ 2 }-6x-1 }{ x+3 } =\underset { h\rightarrow { o }^{ + } }{ lim } \frac { \left( -3-h \right) ^{ 2 }-6(-3-h)-1 }{ -3-h+3 } \)
= -∞
ஃx = -3 is the vertical asymptote,
Also

ஃ y = x - 9 is the slanting asymptote,

\(f(x)=\frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } \) 
\(\underset { x\rightarrow \infty }{ lim } \frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 3 }{ \sqrt { 1+\frac { 2 }{ { x }^{ 2 } } } } \) 
= \(\frac { 3 }{ \sqrt { 1+0 } } =3\) 
ஃy = 3 is the horizontal asymptote.
Also \(\underset { x\rightarrow -\infty }{ lim } \frac { 3x }{ \sqrt { { x }^{ 2 }+2 } } =\underset { \frac { 1 }{ x } \rightarrow { 0 }^{ - } }{ lim } \frac { 3 }{ \sqrt { 1+\frac { 2 }{ { x }^{ 2 } } } } =-3\) 
ஃ y = -3 is the horizontal asymptote

Given \(f(x)=\frac { { x }^{ 2 } }{ { x }^{ 2 }+1 } \)
\(\underset { x\rightarrow { 1 }^{ - } }{ lim } \frac { { x }^{ 2 } }{ x+1 } =\underset { h\rightarrow { o }^{ + } }{ lim } \frac { \left( -1-h \right) ^{ 2 } }{ -1-h+1 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (1+h)^{ 2 } }{ -h } =-\infty \) 
\(\underset { x\rightarrow -1^{ + } }{ lim } \frac { { x }^{ 2 } }{ x+1 } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { (-1+h)^{ 2 } }{ -1+h+1 } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (h-1)^{ 2 } }{ h } =\infty \) 
ஃ x = -1 is the vertical asymptote,
Since the numerator's degree is more than the denominator's degree, let us divide x² by x + 1

ஃy = x-1 is the slanting asymptote,

Given \(f(x)=\frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } \)
\(\quad \underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { x\rightarrow 1^{ + } }{ lim } \frac { { x }^{ 2 } }{ \left( x+1 \right) \left( x-1 \right) } \) 
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1+h \right) ^{ 2 } }{ \left( 1+h+1 \right) \left( 1+h-1 \right) } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { (1+h)^{ 2 } }{ (2+h)(h) } \infty \) 
Also  \(\underset { x\rightarrow 1^{ - } }{ lim } \frac { { x }^{ 2 } }{ (x+1)/(x-1) } =\underset { h\rightarrow 0^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ \left( 1-h+1 \right) \left( 1-h-1 \right) } \)
= \(\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( 1-h \right) ^{ 2 } }{ (2-h)(-h) } =-\infty \) 
ஃ x = -1 and x = 1are vertical asymptotes,
Also
\(\underset { x-\rightarrow \infty }{ lim } \frac { { x }^{ 2 } }{ { x }^{ 2 }-1 } =\underset { \frac { 1 }{ x } \rightarrow 0 }{ lim } \frac { 1 }{ 1-\frac { 1 }{ { x }^{ 2 } } } =1\)
[Divide numerator and denominator by x²]
ஃ y = 1 is a horizontal asymptote,

Let r be the radius of the ripple and A be the area of the ripple.
GIven \(\frac { dr }{ dt } \) = 2 cm/sec and r = 5 cm ... (1)
We know A = πr²
Differentiating with respect to 't' we get,
\(\frac { dA }{ dt } \) = π(2r).\(\frac { dr }{ dt } \)
= π(2) (5) (2) [using (1)]
\(\frac { dA }{ dt } \) = 20 πsq.cm/sec.

Given v = x³
Differentiating with respect to x we get,
\(\frac { dv }{ dt } \) = 3x²
When x = 5, \(\frac { dv }{ dt } \) = 3(52) = 75
∴ \(\frac { dv }{ dt } \) when x = 5 is 75 units.