Page 22 - mathematics

Case (2) Example

Differentiating exponential functions where variables Find dy , where y  x ln x
dx
appear in both the base and exponent:
Solution
y fg ln y  ln f g 
We first take the logarithm of both sides:

ln y  g  ln f  ln y  ln x ln x ln y  ln x ln x

ln y  ln x  2

Next, differentiate both sides with respect to x:

Now: differentiating simple product. y  2 ln x   1 y   2 y ln x   1
y x x

Example Example

 Find dy cosh x dy y x  tan1 xe sinx 3
dx , where y  sin 1 x Find dx , where

Solution Solution

We first take the logarithm of both sides: x ln y  ln tan1 x  sin x 3

 ln y  ln sin1 x cosh x   xy  1  1  3x 2
y  ln y   tan1   1 x   cos x 3
 ln y  cosh x  ln sin1 x

Next, differentiate both sides with respect to x: x 2

y  1 1 
 sin 1  
 y  
 cosh x  x  2  sinh x  ln sin1 x y  1  1  
x  tan1 x   x  
1 x   y
cosh x    2  cos x 3 3x 2  ln y
  1 1   1
 y   y  sin 1 1x 2  
  x   sinh x  ln sin1 x



Taylor Series

* This section shows how functions that are infinitely
differentiable generate power series called Taylor series.

Taylor series generated by f at x = a is:

f x f a  f a   x a f a   x  a2  f a  x  a 3  ....

1! 2! 3!

Maclaurin series generated by f is:

f x f 0  f 0 x f 0  x 2  f 0 x 3  ....

1! 2! 3!

which is the Taylor series generated by f at x = 0.

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