Also if the solution can be on paper it would be easier to follow along thanks in advance.

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.

by Top Rated User (1.1m points)