#### To determine

**To find:** The derivative of the function
y=sin(t+cost).

#### Answer

The derivative of
y=sin(t+cost) is
y′=cos(t+cost)(1−sint2t).

#### Explanation

**Given:**

The function is
y=sin(t+cost).

**Result used:**

**The Chain Rule:**

If *h* is differntibale at *x* and *g* is differentiable at
h(x), then the composite function
F=g∘h defined by
F(x)=g(h(x)) is differentiable at *x* and
F′ is given by the product,

F′(x)=g′(h(x))⋅h′(x) (1)

**Derivative Rule:**

(1) Power Rule:
ddx(xn)=nxn−1.

(2) Product Rule:
ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)]

**Calculation:**

Obtain the derivative of
y.

y′=ddt(y)=ddt(sin(t+cost))

Let
h(t)=t+cost and
g(u)=sinu where u=h(t).

Apply the chain rule as shown in equation (1),

y′(x)=g′(h(x))⋅h′(x) (2)

The derivative
g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=ddu(sinu)=cosu

Substitute
u=t+cost,

g′(h(x))=cos(t+cost)

Thus, the derivative is
g′(h(x))=cos(t+cost).

The derivative of
h(t) is computed as follows,

h′(t)=ddt(t+cost)=ddt(t)+ddt(cost)=(1)+(−sint)ddt(t12)=1+(−sint)12t12−1

Simplify further as,

h′(t)=1+(−sint)12t12−1=1−sint12t

Thus, the derivative of
h(t) is
h′(t)=1−sint12t.

Substitute
cos(t+cost) for
g′(h(x)) and
1−sint12t for
h′(x) in equation (2),

g′(h(x))⋅h′(x)=cos(t+cost)(1−sint12t)=y′=cos(t+cost)(1−sint2t)

Therefore, the derivative of
y=sin(t+cost) is
y′=cos(t+cost)(1−sint2t).