To determine

To find: The differentiation of y=tsint1+t.

Answer

The differentiation of y=tsint1+t is (t+t2)cost+sint(1+t)2_.

Explanation

Given:

The function is y=tsint1+t.

Formula used:

Quotient rule:

If f(t) and g(t) are differentiable functions, then the quotient rule is,

ddt[f(t)g(t)]=g(t)ddt[f(t)]−f(t)ddt[g(t)][g(t)]2 (1)

Product Rule:

If f(t) and g(t) are both differentiable functions, then the product rule is,

ddt[f(t)g(t)]=f(t)⋅ddt[g(t)]+g(t)⋅ddt[f(t)] (2)

Calculation:

Apply the quotient rule as shown in equation (1).

Substitute tsint for f(t) and 1+t for g(t) in equation (1).

ddt[tsint1+t]=1+tddt[tsint]−tsintddt[1+t][1+t]2

Apply the product rule as shown in equation (2).

ddt[tsint1+t]=1+t[tddt[sint]+sintddt[t]]−tsintddt[1+t][1+t]2=1+t[t(cost)+sint(1)]−tsint(0+1)(1+t)2=1+t(tcost+sint)−tsint(1+t)2                 =tcost+sint+t2cost+tsint−tsint(1+t)2

Simplify the equation.

ddt[tsint1+t]=tcost+sint+t2cost(1+t)2=(t+t2)cost+sint(1+t)2

Therefore, the differentiation of y=tsint1+t is (t+t2)cost+sint(1+t)2_.