If y = 2 sin x + 3 cos x, show that: 
Basic idea:
√Second order derivative is nothing but derivative of derivative i.e. ![]()
√The idea of chain rule of differentiation: If f is any real-valued function which is the composition of two functions u and v, i.e. f = v(u(x)). For the sake of simplicity just assume t = u(x)
Then f = v(t). By chain rule, we can write the derivative of f w.r.t to x as:
![]()
√Product rule of differentiation- ![]()
√Apart from these remember the derivatives of some important functions like exponential, logarithmic, trigonometric etc..
Let’s solve now:
Given, y = 2sin x+ 3cos x …..equation 1
As we have to prove :
.
We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.
Let’s find ![]()
As ![]()
So lets first find dy/dx and differentiate it again.
∴ ![]()
[∵
sin x) = cosx &
]
![]()
∴ ![]()
Differentiating again with respect to x :
![]()
![]()
From equation 1 we have :
y = 2 sin x + 3 cos x
∴ ![]()
∴ ![]()
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