Differentiation

Let $f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 1}} + \,\,....\,\, + {a_{n - 1}}x + {a_n}$

$f'(x) = {a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,.....\,\, + {a_{n - 1}}$

$f''(x) = {a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}$

Now,

$f(3x) = {3^n}{a_0}{x^n} + {3^{n - 1}}{a_1}{x^{n - 1}} + {3^{n - 2}}{a_2}{x^{n - 2}} + \,\,....\,\, + 3{a_{n - 1}} + {a_n}$

$f'(x)\,.\,f''(x) = [{a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,....\,\, + {a_{n - 1}}]$

$[{a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}]$

Comparing highest powers of x, we get

${3^n}{a_0}{x_n} = a_0^2(n - 1){x^{n - 1 + n - 2}} = a_0^2{n^2}(n - 1){x^{2n - 3}}$

Therefore, $2n - 3 = n$

$\Rightarrow$ n = 3 and ${3^n}{a_0} = a_0^2{n^2}(n - 1)$

$ \Rightarrow {a_0} = 27 = {3 \over 2}$

Therefore, $f(x) = {3 \over 2}{x^3} + {a_1}{x^2} + {a_2}x + {a_3}$

$f'(x) = {9 \over 2}{x^2} + 2{a_1}x + {a_2}$

$f''(x) = 9x + 2{a_1}$

$f(3x) = {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$

Now, $f(3x) = f'(x)\,.\,f''(x)$

$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$

$ \Rightarrow \left( {{9 \over 2}{x^2} + 2{a_1}x + {a_2}} \right)(9x + 2{a_1})$

$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3} = {{81} \over 2}{x^3} + [9{a_1} + 18{a_1}]{x^2} + [4a_1^2 + 9{a_2}]x + 2{a_1}{a_2}$

Comparing the coefficients, we get

$9{a_1} = 27{a_1}$

$ \Rightarrow {a_1} = 0,\,3{a_2} = 4a_1^2 + 9{a_2} = 9{a_2}$

$ \Rightarrow {a_2} = 0$

Therefore, $f(x) = {3 \over 2}{x^3}$

$f'(x) = {9 \over 2}{x^2}$

$f''(x) = 9x$

Hence, $f''(2) - f'(x) = 18 - 18 = 0$

The given equation is

$y = {({x^2} + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}}$

Differentiating w.r.t. x, we get

${{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) + 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {1 - {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$

Here, we have used the standard differentiatials

${d \over {dx}}{x^n} = n\,{x^{n - 1}}$

That is, ${d \over {dx}}(\sqrt {f(x)} ) = {1 \over {2\sqrt {f(x)} }} \times {d \over {dx}}(f(x))$

Therefore

${{dy} \over {dx}} = {{15{{(x + \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} + {{15{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow \sqrt {{x^2} - 1} {{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{15}} - 15{(x - \sqrt {{x^2} - 1} )^{15}}$

Differentiating w.r.t. x, we get

${{1(2x)} \over {2\sqrt {{x^2} + 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 15 \times 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) - 15 \times 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {{{1 - 1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$

$ \Rightarrow {x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 225{(x + \sqrt {{x^2} - 1} )^{14}}{{(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} - {{225{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow \sqrt {{x^2} - 1} \left[ {{x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}}} \right] = 225{(x + \sqrt {{x^2} - 1} )^{15}} + 225{(x - \sqrt {{x^2} - 1} )^{15}}$

$ \Rightarrow x{{dy} \over {dx}} + ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 225\left[ {{{(x + \sqrt {{x^2} - 1} )}^{15}} + {{(x - \sqrt {{x^2} - 1} )}^{15}}} \right]$

Substituting ${(x + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}} = y$, we get

$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y$

Given,

F(1) = 0, F(3) = 4

F'(x) < 0 for all x $\in$(1, 3)

and f(x) = xF(x)

Now, f'(x) = F(x) + xF'(x)

$\Rightarrow$ f'(1) = F(1) + 1 . F'(1)

$\Rightarrow$ f'(1) = 0 + F'(1) ....... (1)

As F'(x) < 0 for all x $\in$ (1, 3)

$\therefore$ F'(1) < 0

From (1), we get

f'(1) = F'(1) < 0

From Lagrange theorem

$F'(2) = {{F(3) - F(1)} \over {3 - 1}}$

$ = {{ - 4 - 0} \over 2}$

= $-$2

As f(x) = xF(x)

$\therefore$ f(3) = 3 . F(3) = 3 . ($-$4) = $-$12

and f(1) = 1 . F(1) = 0

$\therefore$ $f'(2) = {{f(3) - f(1)} \over {3 - 1}}$

$ = {{ - 12} \over 2} = - 6$

As, f'(x) = F(x) + x . F'(x)

$\therefore$ f'(2) = F(2) + 2 . F'(2)

$ \Rightarrow - 6 = {{f(2)} \over 2} + 2.\,( - 2)$

$ \Rightarrow {{f(2)} \over 2} = - 2$

$\Rightarrow$ f(2) = $-$4

$\therefore$ f(2) < 0

$f'(x) = {{f(3) - f(1)} \over {3 - 1}}$ when x $\in$ (1, 3)

f(3) = $-$12

and f(1) = 0

$\therefore$ $f'(x) = {{ - 12} \over 2} = - 6$

$\therefore$ f'(x) $\ne$ 0 for any x $\in$(1, 3)

Given, $\mathrm{F}^{\prime}(x)=f^{\prime}(x)\quad \text{.... (i)}$

And $\quad \mathrm{F}(x)=\int_0^{x^2} f(\sqrt{t}) d t$

Differentiate the equation (ii) w.r.t. $x$

$\begin{aligned} & \Rightarrow \quad \mathrm{F}^{\prime}(x)=\frac{d\left(x^2\right)}{d x} \cdot f\left(\sqrt{x^2}\right)-\frac{d(0)}{d x} \cdot f(\sqrt{0}) \\ & \Rightarrow \quad f^{\prime}(x)=2 x \cdot f(x)-0 \\ & \Rightarrow \quad \frac{d f(x)}{f(x)}=2 x d x \\ & \Rightarrow \int \frac{d f(x)}{f(x)}=\int 2 x d x \\ & \Rightarrow \ln f(x)=2 \cdot \frac{x^2}{2}+\mathrm{C} \\ & \Rightarrow \ln f(x)=x^2+\mathrm{C} \quad \text{... (iii)} \end{aligned}$

Given, $f(0)=1$

$\begin{array}{llrl} & \therefore & \ln (1) & =0^2+C \\ & \Rightarrow & 0 & =0+C \\ \Rightarrow & & C=0 \end{array}$

Put $c=0$ in the equation (iii)

$\begin{array}{ll} \Rightarrow & \ln f(x)=x^2 \\ \Rightarrow \quad f(x) & =e^{x^2} \\ \Rightarrow \quad f(\sqrt{t}) & =e^t \end{array}$

Put $f(\sqrt{t})=e^t$ in the equation (ii)

$\begin{array}{ll} \Rightarrow & \mathrm{F}(x)=\int_0^{x^2} e^t \cdot d t \\ \Rightarrow & \mathrm{F}(x)=\left[e^t\right]_0^{x^2} \\ \Rightarrow & \mathrm{F}(x)=e^{x^2}-1 \end{array}$

Put $\quad x=2$

$\Rightarrow \mathrm{F}(2)=e^4-1$

Hint:

(i) If $y=\int_{\phi(x)}^{\psi(x)} f(t) d t$, then $\frac{d y}{d x}=\frac{d \psi(x)}{d x} \cdot f(\psi(x))-\frac{d \phi(x)}{d x} \cdot f(\phi(x))$

(ii) $\int \frac{f^{\prime}(x) d x}{f(x)}=\ln (f(x))+c$

Given that,

$g(x) = \log f(x)$

$ \Rightarrow g(x + 1) = \log f(x + 1)$

$ \Rightarrow g(x + 1) = \log xf(x)$

[$\because$ $f(x + 1) = xf(x)$]

$g(x + 1) = \log x + \log f(n)$

$ \Rightarrow g(x + 1) - g(x) = \log (x)$

$g'(x + 1) - g'(x) = {1 \over x}$

$ \Rightarrow g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}$

Putting, $x = x - {1 \over 2}$, we get

$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) = - {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$

Putting $x = 1,2,3$ .........., N we get

$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) - = {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$

Putting $x = 1,2,3$, .........., N we get

$g''\left( {{3 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$ .... (i)

$g''\left( {{5 \over 2}} \right) - g''\left( {{3 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$ ..... (ii)

$g''\left( {{7 \over 2}} \right) - g''\left( {{5 \over 2}} \right) = {{ - {2^2}} \over {{5^2}}}$ ..... (iii)

$g''\left( {N + {1 \over 2}} \right) - g''\left( {N - {1 \over 2}} \right) = {{ - {2^2}} \over {{{(2N - 1)}^2}}}$ ..... (iv)

Adding all the above equations, we get

$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = 4$

$\left[ {1 + {1 \over {{3^2}}} + {1 \over {{5^2}}}\, + \,...\, + \,{1 \over {{{(2N - 1)}^2}}}} \right]$

$f(x) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}}$

for differentiation, we can write f(x) as

$f(x) = 1 - {{2ax} \over {{x^2} + ax + 1}}$

Now, on differentiation

$f'(x) = {{2a({x^2} - 1)} \over {{{({x^2} + ax + 1)}^2}}}$ ..... (i)

$f'(1) = 0 = f'( - 1)$

Then again differentiation (1)

${({x^2} + ax + 1)^2}\,.\,2x - ({x^2} - 1)$

$f''(x) = 2a.\,{{2({x^2} + ax + 1)(2x + a)} \over {{{({x^2} + ax + 1)}^4}}}$

$f''( - 1) = {{ - 4a} \over {{{(2 - a)}^2}}}$

$f''(1) = {{ - 4a} \over {{{(2 - a)}^2}}}$

Combining both

$f''(1){(2 + a)^2} + f''( - 1){(2 - a)^2} = 0$

$\mathop {\lim }\limits_{x \to 0} {{g(x)\cos x - g(0)} \over {\sin x}}$

$ = \mathop {\lim }\limits_{x \to 0} {{g'(x)\cos x - g(x)\sin x} \over {\cos x}}$ (Applying L-Hospital rule)

$g'(0) - 0 = 0 = f''(0)$

Statement - 1 is True.

$f'(x) = g(x)\cos x + g'(x)\sin x$

$f'(0) = g(0)$

Statement - 2 is True.

We have

${y^3} - 3y + x = 0$

Differentiate both sides we get

$3{y^2}.y' - 3y' + 1 = 0$ ..... (i)

Put $y = 2\sqrt 2 ,x = - 10\sqrt 2 $ then

$y'( - 10\sqrt 2 ) = {{ - 1} \over {21}}$

Differentiate eq. (i), we get

$3{y^2}y'' + 6y{(y')^2} - 3y'' = 0$

Put $y = 2\sqrt 2 ,x = - 10\sqrt 2 ,y' = {{ - 1} \over {21}}$ then

$y''( - 10\sqrt 2 ) = {{ - 4\sqrt 2 } \over {{7^3}\,.\,{3^2}}}$

Let’s start by finding the relationship between derivatives.

We know that $\frac{dy}{dx} = \left( \frac{dx}{dy} \right)^{-1}$. This just means that the derivative going from $y$ to $x$ is the reciprocal of going from $x$ to $y$.

To find the second derivative, we get the derivative of the first derivative. So, $\frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d}{dx}\left( \frac{dx}{dy} \right)^{-1}$.

Instead of $x$, let’s write the second derivative in terms of $y$. We use the chain rule for derivatives: $\frac{d}{dx} = \frac{d}{dy} \cdot \frac{dy}{dx}$.

So, $\frac{d^2y}{dx^2} = \frac{d}{dy}\left( \frac{dx}{dy} \right)^{-1} \cdot \frac{dy}{dx}$.

Taking the derivative of $\left( \frac{dx}{dy} \right)^{-1}$ with respect to $y$ gives us $-\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d^2x}{dy^2}$.

Including the extra $\frac{dy}{dx}$ from the chain rule, we get: $\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \frac{d^2x}{dy^2} \cdot \frac{dy}{dx}$.

Now, solve this for $\frac{d^2x}{dy^2}$:

$\frac{d^2x}{dy^2} = \frac{ - d^2y }{ dx^2 } \left( \frac{dy}{dx} \right)^{-3}$.

The statement "The derivative of an even function is always an odd function" is Option A: TRUE.

To understand why this is true, let's first define even and odd functions:

An even function is a function $ f(x) $ that satisfies the condition:

$ f(-x) = f(x) $

An odd function is a function $ g(x) $ that satisfies the condition:

$ g(-x) = -g(x) $

Now, let's consider the derivative of an even function. Let $ f(x) $ be an even function, meaning:

$ f(-x) = f(x) $

We want to find the derivative $ f'(x) $ and determine if it is odd. Using the definition of the derivative, we have:

$ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} $

Since $ f(x) $ is even, we have:

$ f(x) = f(-x) $

Taking the derivative of both sides with respect to $ x $, we get:

$ f'(-x) \cdot (-1) = f'(x) $

or

$ f'(-x) = -f'(x) $

This shows that the derivative $ f'(x) $ satisfies the condition for being an odd function. Therefore, the derivative of an even function is indeed always an odd function.