(a) Consider y=x41.Let x=6116 and Δx=811
Then,Δy=(x+Δx)1/4-x1/4
= (8117)41- (8116)41
= (8117)41 -32
∴(8117)41= 32+Δy
Now,dy is approximately equal to ∆y and is given by,
dy=(dxdy) Δx= 4(x)431(Δx) (as y=x41)
= 4(8116)431(811)= 4×827x 811=32×31=961=0.010
Hence, the approximate value of 811741 is 32+0.010=0.667+0.010=0.677.
(b)Consider y=x=51 . Let x=32 and ∆x=1.
Δy=(x+Δx)−51−x−51=33− (51) −(32)− (51) − (21)
∴(33)− (51)=21+Δy
Now,dy is approximately equal to ∆y and is given by,
dy=(dxdy)(Δx)=− 5 (x)(56)1 (Δx) ( as y=51)
=-5(2)61(1)=−3201=−0.003
Hence, the approximate value of(33)5−1 is 21+(-0.003)
=0.5−0.003=0.497.