If x =a (cos θ + θ sin θ), y=a (sin θ – θ cos θ) prove that

(sin θ + θ cosθ) and .

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..


The idea of parametric form of differentiation:


If y = f (θ) and x = g(θ), i.e. y is a function of θ and x is also some other function of θ.


Then dy/dθ = f’(θ) and dx/dθ = g’(θ)


We can write :


Given,


x =a (cos θ + θ sin θ) ……equation 1


y =a (sin θ – θ cos θ) ……equation 2


to prove :



(sin θ + θ cosθ)


.


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




[ differentiated using product rule for θsinθ ]


..eqn 4


Again differentiating w.r.t θ using product rule:-




Similarly,




………….equation 5


Again differentiating w.r.t θ using product rule:-





Using equation 4 and 5 :



As


again differentiating w.r.t x :-



[using chain rule]



Putting a value in the above equation-


We have :




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