 Hi and welcome to the session. I am Deepika here. Let's discuss a question. If x is equal to a cos t plus t sin t and y is equal to a sin t minus t cos t find d2y by dx square. So let's start the solution given x is equal to a cos t plus t sin t on differentiating with respect to t we get we have dx by dt is equal to a Now derivative of cos t is minus sin t plus for this portion, let's apply the product rule t into cos t plus sin t into 1 So this is equal to dx by dt is equal to a t cos t Now y is given to us y is equal to a sin t minus t cos t Again differentiating with respect to t dy by dt is equal to a derivative of sin t is cos t again for this portion let's apply the product rule we get minus now t into derivative of cos t that is minus sin t plus cos t into into 1 therefore dy by dt is equal to a cos t plus t sin t minus cos t We have dy by dt is equal to a t sin t Now by chain rule we have dy by dx is equal to dy by dt Into dt by dx dy by dt is equal to a t sin t so this is equal to a t sin t Into dt by dx that is 1 over a t cos t So this is equal to we get tan t So this is our dy by dx Now again differentiating with respect to x we have d2 y by dx square Now derivative of tan t is secant square t Into derivative of t with respect to x that is dt by dx so this is equal to therefore d2 y by dx square is equal to secant square t Into dt by dx is 1 over a t cos t So this is equal to qt upon a t where t is between 0 and pi by 2 So hence we have Find d2 y by dx square and our answer is secant cube t upon a t 0 is less than t t is less than pi by 2 I Hope the question is clear to you by and have a nice day