Let's find the derivative of the given function f(x)=x3sin(x2)+ln(x)cos(ex).
We will use the sum rule, product rule, and chain rule for differentiation.
The sum rule states that dxd[g(x)+h(x)]=dxd[g(x)]+dxd[h(x)].
The product rule states that dxd[u(x)v(x)]=u′(x)v(x)+u(x)v′(x).
The chain rule states that dxd[f(g(x))]=f′(g(x))g′(x).
Step 1: Differentiate the first term, x3sin(x2).
Let u(x)=x3 and v(x)=sin(x2).
Then u′(x)=dxd[x3]=3x2.
To find v′(x), we apply the chain rule. Let w=x2, so v(x)=sin(w).
Then dwdv=cos(w) and dxdw=2x.
So, v′(x)=dxd[sin(x2)]=cos(x2)⋅(2x)=2xcos(x2).
Using the product rule for the first term:
dxd[x3sin(x2)]=u′(x)v(x)+u(x)v′(x)
=(3x2)sin(x2)+(x3)(2xcos(x2))
=3x2sin(x2)+2x4cos(x2)
Step 2: Differentiate the second term, ln(x)cos(ex).
Let p(x)=ln(x) and q(x)=cos(ex).
Then p′(x)=dxd[ln(x)]=x1.
To find q′(x), we apply the chain rule. Let z=ex, so q(x)=cos(z).
Then dzdq=−sin(z) and dxdz=ex.
So, q′(x)=dxd[cos(ex)]=−sin(ex)⋅(ex)=−exsin(ex).
Using the product rule for the second term:
dxd[ln(x)cos(ex)]=p′(x)q(x)+p(x)q′(x)
=(x1)cos(ex)+(ln(x))(−exsin(ex))
=xcos(ex)−exln(x)sin(ex)
Step 3: Combine the derivatives of both terms.
f′(x)=dxd[x3sin(x2)]+dxd[ln(x)cos(ex)]
f′(x)=(3x2sin(x2)+2x4cos(x2))+(xcos(ex)−exln(x)sin(ex))
f′(x)=3x2sin(x2)+2x4cos(x2)+xcos(ex)−exln(x)sin(ex)
The final answer is f′(x)=3x2sin(x2)+2x4cos(x2)+xcos(ex)−exln(x)sin(ex).