Differentiation (Part-38) | S N De

 

Some facts about "Differentiation" :

In mathematics, differentiation is a process used to find the rate at which a function changes. It involves finding the derivative of a function, which gives the slope of the function at any given point.

The derivative of a function can be thought of as the instantaneous rate of change of the function at a particular point. It is defined as the limit of the ratio of the change in the value of the function to the change in the input variable as the change in the input variable approaches zero.

Differentiation is an important tool in many branches of mathematics, including calculus, optimization, and physics. It is used to solve problems involving rates of change, optimization, and related rates.

The power rule is a formula used to differentiate polynomials. It states that the derivative of x to the power of n is equal to n times x to the power of n-1.

The chain rule is a formula used to differentiate composite functions. It states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.

In calculus, the second derivative of a function is the derivative of its derivative. It gives information about the rate at which the slope of the function is changing.

Differentiation is closely related to integration, which is the reverse process of finding a function given its derivative. Together, differentiation and integration make up the branch of mathematics known as calculus.

Now, we are going to solve few problems in S N Dey Mathematics ( Class-12 ).

12. If $~~f(x)=\frac 1x,~$ then find the value of $\, \theta\,$ in the mean value theorem $~~f(x+h)=f(x)+hf'(x+\theta h),~0 <\theta<1.$

Sol. We have, $~f(x)=\frac 1x=x^{-1} \rightarrow(1), \\ \therefore~f'(x)=-x^{-1-1}=-\frac{1}{x^2}\rightarrow(2)$

Now, from the mean value theorem we get,

$~~f(x+h)=f(x)+hf'(x+\theta h) \\ \text{or,}~~\frac{1}{x+h}=\frac 1x+h~\cdot\left[-\frac{1}{(x+\theta h)^2}\right]~~[\text{By (1),(2)}] \\ \text{or,}~~ \frac{1}{x+h}-\frac 1x=-\frac{h}{(x+ \theta h)^2} \\ \text{or,}~~ \frac{x-(x+h)}{x(x+h)}=-\frac{h}{(x+\theta h)^2} \\ \text{or,}~~ \frac{x-x-h}{x(x+h)}=-\frac{h}{(x+\theta h)^2} \\ \text{or,}~~ \frac{-h}{x(x+h)}=\frac{-h}{(x+\theta h)^2}~~[\because~ h\neq 0] \\ \therefore~~x(x+h)=(x+\theta h)^2 \\ \text{or,}~~ x^2+xh=x^2+2xh \theta+\theta^2h^2 \\\text{or,}~~  xh=2xh \theta+\theta^2h^2 \\ \text{or,}~~ \theta^2h^2+2xh\theta-xh=0 \\ \text{or,}~~ \theta =\frac{-2xh \pm \sqrt{(2xh)^2-4(-xh)\cdot h^2}}{2h^2} \\ \text{or,}~~ \theta=\frac{-2xh \pm \sqrt{4x^2h^2+4xh \cdot h^2}}{2h^2} \\ \text{or,}~~ \theta=\frac{-2xh \pm \sqrt{4h^2(x^2+xh)}}{2h^2}\\ \text{or,}~~ \theta=\frac{-2xh \pm 2h\sqrt{x^2+xh}}{2h^2} \\ \text{or,}~~ \theta=\frac{2h(-x \pm \sqrt{x^2+xh})}{2h \cdot h}\\ \text{or,}~~ \theta=\frac{\sqrt{x^2+xh}-x}{h}~~\text{(ans.)}$

13. Using Lagrange's mean value theorem, find a point on the curve $~~y=\sqrt{x-2}~$ defined in the interval $~[2,3]~$ where the tangent to the curve is parallel to the chord joining the end points of the curve.

Sol. let $~~y=f(x)=\sqrt{x-2}\rightarrow(1) \\ \therefore f'(x)\\=\frac{d}{dx}(x-2)^{1/2}\\=\frac 12(x-2)^{\frac 12-1}\\=\frac{1}{2\sqrt{x-2}}\rightarrow(2)$ 

Now, for $~f(x)~$ to be defined in $~[2,3]~$ we must have $~x-2 \geq 0~~$ which means $~~x \geq 2.$

Clearly, $~f(x)~$ is continuous in $~[2,3]~,$ differentiable in $~(2,3)~.$ So, according to Lagrange's mean value theorem,  there exists at least one value of $~x~$ say $~x=c \in (2,3)~$ such that 

$~~f'(c)=\frac{f(3)-f(2)}{3-2} \\ \text{or,}~~\frac{1}{2\sqrt{c-2}}=\frac{\sqrt{3-2}-\sqrt{2-2}}{1}~~~[\text{By (1),(2)}] \\ \text{or,}~~\frac{1}{2\sqrt{c-2}}=\sqrt{1}-0=1 \\ \text{or,}~~\left(\frac{1}{2\sqrt{c-2}}\right)^2 =1^2 \\ \text{or,}~~ \frac{1}{4(c-2)}=1 \\ \text{or,}~~4c-8=1 \\ \text{or,}~~ 4c=1+8 \\ \text{or,}~~ 4c=9 \\ \text{or,}~~ c=\frac 94 $

Now, for $~c=\frac 94~,$ the corresponding value of 

$~~~f(c)\\=\sqrt{\frac 94-2}\\=\sqrt{\frac{9-8}{4}}\\=\sqrt{\frac 14}\\=\frac 12.$

Hence, the point on the curve $~~y=\sqrt{x-2}~$ defined in the interval $~[2,3]~$ where the tangent to the curve is parallel to the chord joining the end points of the curve is  $~~~(\frac 94,\frac 12).$

Elements of Mathematics For Class XII (Vol-I and Vol-II) Paperback

14. Using Lagrange's mean value theorem, find a point on the curve $~~y=x^2~$ where the tangent to the curve is parallel to the line joining the points $~(1,1)~$ and $~(2,4).$

Sol. Here the given curve $~y=x^2=f(x)~\text{(say)}\rightarrow(1) \\ \therefore~f'(x)=2x\rightarrow(2)$ 

Clearly, $~f(x)~$ is a polynomial function which is continuous and differentiable everywhere in a finite interval. In this problem, we consider the interval $~[1,2].$ 

Hence, Lagrange's mean value theorem is applicable here. So, there exists a point $~c \in (1,2)~$ such that 

$~~~f'(c)=\frac{f(2)-f(1)}{2-1}\\ \text{or,}~~2c=\frac{2^2-1^2}{1}~~[\text{By (1),(2)}] \\ \text{or,}~~2c=4-1 \\ \text{or,}~~2c=3 \\ \text{or,}~~ c=\frac 32.$

Finally, $~~f(c)=c^2 \Rightarrow f(3/2)=(3/2)^2=\frac{9}{4}.$

So, $~~(c,f(c))=\left(\frac 32, \frac 94 \right)~$ is the required point on the curve $~~y=x^2~$ where the tangent to the curve is parallel to the line joining the points $~(1,1)~$ and $~(2,4).$