To determine
To find: The derivative of d35dx35(xsinx).
Answer
The derivative of d35dx35(xsinx) is −xcosx−35sinx_.
Explanation
Derivative rules used:
(1) Product Rule:ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)]
(2) Sum Rule :ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]
(3) Constant Multiple Rule :ddx[cf(x)]=c⋅ddx[f(x)]
Calculation:
Let f(x)=xsinx (1)
Substitute h(x)=sinx in equation (1),
Obtain the first derivative of f(x).
f′(x)=ddx[xh(x)]
Apply the product rule (1) and simplify further,
f′(x)=xddx(h(x))+h(x)ddx(x)=xh′(x)+h(x)(1)=xh′(x)+h(x)
Thus, the first derivative of f(x) is f′(x)=xh′(x)+h(x).
Obtain the second derivative of f(x).
f″(x)=d2dx2(f(x))=ddx(f′(x))=ddx(xh′(x)+h(x))
Apply the derivative rules (1), (2) and (3),
f″(x)=ddx(xh′(x))+ddx(h(x))=xddx(h′(x))+h′(x)ddx(x)+ddxh(x)=x(h″(x))+h′(x)(1)+h′(x)=xh″(x)+2h′(x)
Thus, the second derivative of f(x) is f″(x)=xh″(x)+2h′(x).
Obtain the third derivative of f(x).
f‴(x)=d3dx3(f(x))=ddx(f″(x))=ddx(xh″(x)+2h′(x))
Apply the derivative rules (1) and (2),
f‴(x)=ddx(xh″(x))+ddx(2h′(x))=[xddx(h″(x))+h″(x)ddx(x)]+2ddx[h′(x)]=x(h‴(x))+h″(x)(1)+2h″(x)=xh‴(x)+3h″(x)
Thus, the third derivative of f(x) is f‴(x)=xh‴(x)+3h″(x).
Proceed in the similar way, the nth derivative of f(x) is f(n)(x)=xh(n)(x)+nh(n−1)(x).
Substitute n=35 and h(x)=sinx in the above equation,
f(35)(x)=xh(35)(x)+35h(35−1)(x)=xh(35)(x)+35h(34)(x)=x(d35dx35(h(x)))+35(d34dx34(h(x)))=x(d35dx35(sinx))+35(d34dx34(sinx))
Since every fourth derivative of sinx is sinx, the derivative f(35)(x) can be expressed as follows,
f(35)(x)=x(d35dx35(sinx))+35(d34dx34(sinx))=x(d32+3dx32+3(sinx))+35(d32+2dx32+2(sinx)) (Q35=4(8)+3 34=4(8)+2)=x(d3dx3(d32dx32(sinx)))+35(d2dx2(d32dx32(sinx)))
Therefore, the equation is d35dx35(f(x))=x(d3dx3(sinx))+35(d2dx2(sinx)) (2)
Obtain the derivative d2dx2(sinx).
d2dx2(sinx)=ddx(ddx(sinx))=ddx(cosx)=−sinx
Thus, d2dx2(sinx)=−sinx.
Obtain the derivative d3dx3(sinx).
d3dx3(sinx)=ddx(d2dx2(sinx))=ddx(−sinx)=−ddx(sinx)=−cosx
Thus, d3dx3(sinx)=−cosx.
Substitute d2dx2(sinx)=−sinx, d3dx3(sinx)=−cosx and f(x)=xsinx in equation (2),
d35dx35(xsinx)=−xcosx−35sinx
Therefore, the derivative of d35dx35(xsinx) is −xcosx−35sinx_.