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Q 15)
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

Given function is Given function is satisfies for the all real values of x case (i)  k < 0 Hence, function is continuous for all values of x < 0 case (ii)  x = 0 L.H.L at x= 0 R.H.L. at x = 0 L.H.L. = R.H.L. = f(0) Hence, function is continuous at x = 0 case (iii)  k > 0 Hence , function is continuous for all values of x > 0 case (iv) k < 1  Hence , function is...

Q23  If  $y = e ^{a \cos ^{-1}x} , -1 \leqx \leq 1$ , show that $( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$

Given function is Now, differentiate w.r.t x we will get                       -(i) Now, again differentiate w.r.t x                                                                             -(ii) Now, we need to show that Put the values from equation (i) and (ii) Hence proved

Q22  If  $y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$ , prove that $dy/dx = \begin{vmatrix} f '(x) & g'(x) & h' (x) \\ l& m &n \\ a& b &c \end{vmatrix}$

Given that We can rewrite it as Now, differentiate w.r.t x we will get Hence proved

Q21  Does there exist a function which is continuous everywhere but not differentiable

Consider f(x) = |x| + |x +1| We know that modulus functions are continuous everywhere and sum of two continuous function is also a continuous function Therefore, our function f(x) is continuous Now, If Lets differentiability of our function at x = 0 and x= -1 L.H.D. at x = 0                                                                                                                ...

Q20 Using the fact that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and the differentiation,
obtain the sum formula for cosines.

Given function is Now, differentiate w.r.t. x                                                        Hence, we get the formula by differentiation of sin(A + B)

Q19  Using mathematical induction prove that $\frac{d}{dx} (x^n) = nx ^{n-1}$ for all positive  integers n.

Given equation is We need to show that     for all positive  integers n Now, For ( n = 1)     Hence, true for n = 1 For (n = k)      Hence, true for n = k For ( n = k+1)                                                                                                    Hence, (n = k+1) is true whenever (n = k) is true Therefore, by the principle of  mathematical induction  we can say...

Q18  If$f (x) = |x|^3$, show that f ''(x) exists for all real x and find it.

Given function is Now, differentiate in both the cases And In both, the cases f ''(x) exist Hence, we can say that f ''(x) exists for all real x and values are

Q17  If   $x = a (\cos t + t \sin t)$ and   $y = a (\sin t - t \cos t),$ find $\frac{d^2 y }{dx^2 }$

Given functions are   and    Now, differentiate both the functions w.r.t. t  independently We get                                                        Similarly,                                                         Now, Now, the second derivative                                                                                                    Therefore,

Q16  If $\cos y = x \cos (a + y)$, with $\cos a \neq \pm 1$ , prove that  $\frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }$

Given  function is Now,  Differentiate w.r.t x Hence proved

Q15  If $(x - a)^2 + (y - b)^2 = c^2$ , for some c > 0, prove that$\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}}\:$ is a constant independent of a and b.

Given function is              - (i) Now, differentiate w.r.t. x       -(ii) Now, the second derivative Now, put values from equation (i) and (ii)                     Now,            Which is independent of a and b  Hence proved

Q14  If $x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}$

Given function is Now, squaring both sides Now, differentiate w.r.t. x is Hence proved

Q 13    Find dy/dx   if $y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0

Given function is  Now, differentiatiate w.r.t. x  Therefore, differentiatiate w.r.t. x  is 0

Q12  Find  dy/dx    if $y = 12 (1 - \cos t), x = 10 (t - \sin t),$ $-\frac{\pi }{2}

Given  equations are Now, differentiate both y and x w.r.t t independently And Now                                                                                                                    Therefore, differentiation w.r.t x is

Q11  $x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$

Given function is  take  Now, take log on both the sides Now, differentiate w.r.t x                    -(i) Similarly, take  Now, take log on both the sides Now, differentiate w.r.t x                -(ii) Now Put the value from equation (i) and (ii) Therefore, differentiation w.r.t x is

Q10    $x ^x + x ^a + a ^x + a ^a$  , for some fixed a > 0 and x > 0

Given function is Lets take  Now, take log on both sides Now, differentiate w.r.t x            -(i) Similarly, take  take log on both the sides Now, differentiate w.r.t x            -(ii) Similarly, take  take log on both the sides Now, differentiate w.r.t x     -(iii) Similarly, take  take log on both the sides Now, differentiate w.r.t x                          ...

Q9  $(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4}

Given function is Take log on both the sides  Now, differentiate w.r.t. x Therefore, differentiation w.r.t x is

Q 8 $\cos ( a \cos x + b \sin x )$, for some constant a and b.

Given function is Now, differentiation w.r.t x                                                                                                                 Therefore, differentiation w.r.t x

Q7  Differentiate w.r.t. x the function in Exercises 1 to 11. $( \log x )^{ \log x } , x > 1$

Given function is Take log on both sides  Now, differentiate w.r.t. Therefore, differentiation w.r.t x is

Q6    Differentiate w.r.t. x the function in Exercises 1 to 11.

$\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$

Given function is Now, rationalize the [] part                                                                                                                                                                                                                                                                                                                                                      ...

Q5  Differentiate w.r.t. x the function in Exercises 1 to 11.

$\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$

Given function is Now, differentiation w.r.t. x is By using the Quotient rule                                                                           Therefore, differentiation w.r.t. x is
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