Page 138 - Foundations Of Differential Calculus

6. On the Differentiation of Transcendental Functions 121
√
1
1
1
Since 2(1 − cos x) = 2 sin x and sin x = 2 sin x cos x,wehave
2
2
2
1 1
dy = dx cos x,
2 2
1
which follows immediately from the form y = sin x.
2
1 1
III. If y = cos ln ,welet p =ln so that y = cos p. Hence
x x
dy = −dp sin p.
But since p =ln 1 − ln x,wehave dp = −dx/x, and so
dx 1
dy = sin ln .
x x
IV. If y = e sin x ,wehave

dy = e sin x dx cos x.

V. If y = e −n/cos x , then

e −n/cos x ndx sin x
dy = − .
2
cos x
√
VI. If y =ln 1 − 1 − e −n/sin x ,welet p = e −n/sin x , and since

y =ln 1 − 1 − p ,
we have
dp
dy = √ √ .
2 1 − 1 − p 1 − p
But

e −n/sin x ndx cos x
dp = 2 .
sin x
When this value is substituted, we obtain
ne −n/sin x dx cos x
dy = √ √ .
2 −n/sin x −n/sin x
2 sin x 1 − 1 − e 1 − e