#### To determine

**To find:** The derivative of
y=sin(1+x2).

#### Answer

The derivative of
y=sin(1+x2) is
xcos(1+x2)sin(1+x2).

#### Explanation

**Derivative Rules:**

**The Chain Rule:**

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

F′(x)=f′(g(x))⋅g′(x) (1)

(1).**Power Rule:**
ddx(xn)=nxn−1

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

**Calculation:**

Obtain the derivative of *y.*

y′=ddx(sin(1+x2))=ddx(sin(1+x2))12

Apply chain rule, power rule and simplify as,

y′=(12)(sin(1+x2))12−1ddx(sin(1+x2))=(sin(1+x2))−122ddx(sin(1+x2))=12sin(1+x2)ddx(sin(1+x2))

Apply chain rule, power rule and simplify as,

y′=12sin(1+x2)(cos(1+x2))ddx(1+x2)

Apply sum rule, power ruke and simplify as,

y′=12sin(1+x2)(cos(1+x2))[ddx(1)+ddx(x2)]=12sin(1+x2)(cos(1+x2))[0+2x2−1]=2x2sin(1+x2)(cos(1+x2))=x(cos(1+x2))sin(1+x2)

Therefore, the derivative of
y=sin(1+x2) is
xcos(1+x2)sin(1+x2).