#### To determine

**To find:** The first the derivative by finding first few derivatives and then observing the pattern that occurs

#### Answer

The derivative of
y=D35xsinπx is
(−1)18(35)π34sinπx+(−1)17xπ35cosπx.

#### Explanation

**Given:**

The function is
y=D35xsinπx.

**Formula used:**

**Chain Rule:**

If *h* is differentiable 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)

(1) Product Rule:
ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

**Calculation:**

Use product rule and obtain the derivative of
f(x)=xsinπx

f1(x)=xddx(sinπx)+sinπxddx(x)=xcosπx(π)+sinπx(1)=xπcosπx+sinπx

Similarly obtain the second derivative of the function as follows.

f2(x)=xddx(πcosπx)+πcosπxddx(x)+ddx(sinπx)=xπ(−sinπx)(π)+πcosπx(1)+cosπx(π)=2πcosπx−xπ2sinπx

Similarly obtain the third derivative of the function as follows.

f3(x)=2πddx(cosπx)−π2xddx(sinπx)−π2sinπxddx(x)=2π(−sinπx)(π)−π2x(cosπx)(π)−π2sinπx(1)=2π2(−sinπx)−π2x(cosπx)(π)−π2sinπx=−3π2sinπx−π3xcosπx

Here arises two cases and the formula can be generalized as follows.

When *n* is even,

f2n(x)=(−1)n+12nπ2n−1cosπx+(−1)n+1xπ2ksinπx (1)

When *n* is odd,

f2n+1(x)=(−1)n+1(2n+1)π2nsinπx+(−1)nxπ2n+1cosπx (2)

Since 35 is odd, substitute
n=17 in the above equation (2) as,

f2(17)+1(x)=(−1)17+1(2(17)+1)π2(17)sinπx+(−1)17xπ2(17)+1cosπx=(−1)18(35)π34sinπx+(−1)17xπ35cosπx

Therefore, the derivative of
y=D35xsinπx is
(−1)18(35)π34sinπx+(−1)17xπ35cosπx