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_.