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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
Given that We can rewrite it as Now, differentiate w.r.t x we will get Hence proved
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                                                                                                                ...
Given function is Now, differentiate w.r.t. x                                                        Hence, we get the formula by differentiation of sin(A + B)                                                                            
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...
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   
Given functions are   and    Now, differentiate both the functions w.r.t. t  independently We get                                                        Similarly,                                                         Now, Now, the second derivative                                                                                                    Therefore,  
Given  function is Now,  Differentiate w.r.t x Hence proved
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
Given function is Now, squaring both sides Now, differentiate w.r.t. x is Hence proved
Given function is  Now, differentiatiate w.r.t. x  Therefore, differentiatiate w.r.t. x  is 0
Given  equations are Now, differentiate both y and x w.r.t t independently And Now                                                                                                                    Therefore, differentiation w.r.t x is 
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    
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                          ...
Given function is Take log on both the sides  Now, differentiate w.r.t. x Therefore, differentiation w.r.t x is 
Given function is Now, differentiation w.r.t x                                                                                                                 Therefore, differentiation w.r.t x  
Given function is Take log on both sides  Now, differentiate w.r.t. Therefore, differentiation w.r.t x is 
Given function is Now, rationalize the [] part                                                                                                                                                                                                                                                                                                                                                      ...
Given function is Now, differentiation w.r.t. x is By using the Quotient rule                                                                           Therefore, differentiation w.r.t. x is   
Given function is Now, differentiation w.r.t. x is                                                                                                                                                                                                                               Therefore, differentiation w.r.t. x is   
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