How do you find the derivative of # (tan^3(x)+10)^2#?

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Answer 1 · 2018-03-04T09:45:08

I get #6tan^2(x)sec^2(x)(tan^3(x)+10)#.

We use the chain rule, which states that

#(df)/dx=(df)/(du)*(du)/dx#

Let #u=tan^3(x)+10#, so we have #f=u^2#, then #(df)/(du)=2u#

We can also differentiate #tan^3(x)# using the chain rule.

We let #z=tan(x),f=z^3#, then #(df)/dz=3z^2#, and #(dz)/dx=sec^2(x)#.

So, the derivative of #tan^3(x)# is #3z^2sec^2(x)=3tan^2(x)sec^2(x)#.

That is also the #(du)/dx#.

Putting everything back together, we get

#(df)/dx=2u*3tan^2(x)sec^2(x)#

Replacing back #u=tan^3(x)+10#, we get

#(df)/dx=2(tan^3(x)+10)*3tan^2(x)sec^2(x)#

#=6tan^2(x)sec^2(x)(tan^3(x)+10)#