We are going to need dx/dt and dy/dt.
dx/dt=(11-e)/2e+1/(1+t²); dy/dt=(2-t)eᵗ-eᵗ-11.
We can find dz/dt in several stages:
dz/dt=(20+x)d(y^tan(x))/dt+y^tan(x)d(20+x)/dt (product rule).
Note that y^tan(x)d(20+x)/dt=y^tan(x)dx/dt.
Let’s take a closer look at d(y^tan(x))/dt, because it contains two variables.
First we need to express it in the form eᵖ. So let p=ln(y^tan(x))=tan(x)ln(y).
dp/dt=tan(x)(1/y)dy/dt+ln(y)sec²(x)dx/dt (product rule).
Let q=eᵖ, dq/dt=eᵖdp/dt=eᵖ(tan(x)(1/y)dy/dt+ln(y)sec²(x)dx/dt).
Therefore d(y^tan(x))/dt=dq/dt=eᵖ(tan(x)(1/y)dy/dt+ln(y)sec²(x)dx/dt).
Rather than expand this fully, we can try to evaluate each term in dz/dt by putting t=1, starting with x and y:
x=((11-e)/2e)(t-1)+tan⁻¹(t)=0+π/4=π/4;
y=(2-t)eᵗ-11(t-1)=e.
Next, the derivatives:
dx/dt=(11-e)/2e+½, dy/dt=e-e-11=-11.
Next, evaluate p=tan(x)ln(y)=tan(π/4)ln(e)=1.
dq/dt=eᵖ(tan(x)(1/y)dy/dt+ln(y)sec²(x)dx/dt),
dq/dt=e(tan(π/4)(1/e)(-11)+ln(e)sec²(π/4)((11-e)/2e+½)).
dq/dt=e(-11/e+2((11-e)/2e+½))=-11+2e((11-e)/2e+½),
dq/dt=-11+11-e+e=0.
dz/dt=(20+x)dq/dt+y^tan(x)dx/dt,
dz/dt=0+e((11-e)/2e+½)=(11-e)/2+e/2=11/2.
The answer is 11/2.