How do you differentiate #f(x)=sec(e^(sqrtx-4) ) # using the chain rule?

1 Answer
mason m
Jan 23, 2016

#f'(x)=(e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4)))/(2sqrtx)#

Explanation:

The chain rule states that #d/dx[sec(u)]=u'*sec(u)tan(u)#, and here #u=e^(sqrtx-4)#.

#f'(x)=d/dx[e^(sqrtx-4)]*sec(e^(sqrtx-4))tan(e^(sqrtx-4))#

To find this derivative, use the rule that #d/dx[e^u]=u'*e^u#, and this time #u=sqrtx-4#.

#f'(x)=d/dx[sqrtx-4]*e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4))#

To find #d/dx[sqrtx-4]#, use the power rule, in that #d/dx[x^(1/2)-4]=1/2x^(-1/2)=1/(2sqrtx)#. This gives

#f'(x)=(e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4)))/(2sqrtx)#

Related questions
See all questions in Chain Rule
Impact of this question
1396 views around the world
You can reuse this answer
Creative Commons License