How do you differentiate #f(x)= tanx# twice using the quotient rule?

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Answer 1 · 2015-12-23T23:51:18

#dy/dx=sec^2x#

#(d^2y)/dx^2=2sec^2xtanx#

In order to start properly, let's just remember that #tanx=(sinx)/(cosx)# and now differentiate this quotient using the proper rule.

The quotient rule states that for a function #y=f(x)/g(x)#, #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2#

So, as for your function:

#f(x)=sinx#
#g(x)=cosx#
#f'(x)=cosx#
#g'(x)=-sinx#

#(dy)/(dx)=(cosxcosx-sinx(-sinx))/(cosx)^2=(cos^2x+sin^2x)/cos^2x#

From trigonometric identities, we know that #sin^2x+cos^2x=1#

#(dy)/(dx)=1/(cos^2x)=sec^2x#

To find the second derivative, use the chain rule, which states that for a function #y=f(g(x))#, #dy/dx=f'(g(x))g'(x)#.

First, rewrite #dy/dx=cos^-2x#

Then, according to the chain rule:

#(d^2y)/dx^2=-2cos^-3xd/dx(cosx)#

#=>(-2(-sinx))/cos^3x#

Your interpretation of a final answer could vary.

#(d^2y)/dx^2=(2sinx)/cos^3x# #"or"# #(d^2y)/dx^2=2sec^2xtanx#