Loading [MathJax]/jax/output/HTML-CSS/jax.js
Back to the class
Section 5.4 #49: Calculate ddxex+1ex−1.
Solution: Using the quotient rule, compute
ddxex+1ex−1=(ex−1)(ex)−(ex+1)(ex)(ex−1)2=ex[(ex−1)−(ex+1)](ex−1)2=−2ex(ex−1)2.◼
Section 5.4 #52: Calculate ddxe2xtan(2x).
Solution: Recall that ddxtan(x)=sec2(x). Therefore use the product rule to compute
ddxe2xtan(2x)=ddx[e2x]tan(2x)+e2xddxtan(2x)=2e2xtan(2x)+2e2xsec2(2x).◼
Section 5.4 #106: Calculate ∫e2x+2ex+1exdx.
Solution: You could make the substitution u=ex, but it is easier to use algebra to simplify the integrand:
e2x+2ex+1ex=ex+2+e−x,
so we may write
∫e2x+2ex+1exdx=∫ex+2+e−xdx=∫exdx+∫2dx+∫e−xdx.
The first two of these integrals are straightforward, and for the third one we make the substitution u=−x so that −du=dx and we compute
∫e2x+2ex+1exdx=∫exdx+∫2dx+∫e−xdx=ex+2x−∫eudu=ex+2x−eu+C=ex+2x−e−x+C.◼
Section 5.4 #107: Calculate ∫e−xtan(e−x)dx.
Solution: First we recall how to compute ∫tan(x)dx: since tan(x)=sin(x)cos(x) we make the substitution u=cos(x) so that −du=sin(x)dx and we see
∫tan(x)dx=∫sin(x)cos(x)dx=−∫1udu=−log(u)+C=−log(cos(x))+C.
Now we may proceed with the original problem. To do it, make the subtitution w=e−x so that −dw=e−xdx (note: I'm using w instead of u since I already used u above). Now compute
∫e−xtan(e−x)dx=−∫tan(w)dw=−[−log(cos(w))]+C=log(cos(e−x))+C.◼
Section 5.5 #43: Calculate ddtt22t.
Solution: Recall that we may write
2t=elog(2t)=etlog(2).
Now use the product rule to compute
ddtt22t=ddt[t2]2t+t2ddtetlog(2)=(2t)(2t)+(t2)(log(2)etlog(2))=t2t[2+tlog(2)].◼
Section 5.5 #81: Calculate ∫105x−3xdx.
Solution: Recall how to compute ∫axdx for any a>0: first rewrite ax=elog(ax)=exlog(a) and use the u-substitution u=xlog(a) to get 1log(a)du=dx. Then,
∫axdx=∫exlog(a)dx=1log(a)∫eudu=1log(a)eu+C=1log(a)exlog(a)+C=1log(a)ax+C.
Applying this formula for a=5 and for a=3 allows us to compute
∫105x−3xdx=∫105xdx−∫103xdx=[1log(5)5x|10−[1log(3)3x|10=5−1log(5)−3−1log(3)=4log(5)−2log(3).◼