Question #336d1

1 Answer
Mar 17, 2017

#d/dx(x^2-tanx) = 2x-1/cos^2x#

Explanation:

By definition the derivative of a function #f(x)# is:

#(df)/dx = lim_(h->0) (f(x+h)-f(x))/h#

Substituting: #f(x) = x^2-tanx#:

#(df)/dx = lim_(h->0) ((x+h)^2-x^2 -tan(x+h)+tanx)/h #

If the limits exist separately we have:

#(df)/dx = lim_(h->0) ((x+h)^2-x^2)/h -lim_(h->0)(tan(x+h)-tanx)/h #

So let's evaluate separately:

#lim_(h->0) ((x+h)^2-x^2)/h = lim_(h->0) (cancel(x^2)+2hx+h^2-cancel(x^2))/h#

#lim_(h->0) ((x+h)^2-x^2)/h = lim_(h->0) cancelh(2x+h)/cancelh#

#lim_(h->0) ((x+h)^2-x^2)/h = 2x#

and using the identity:

#tan(a+b) = (tana + tanb)/(1-tanatanb)#

#lim_(h->0) (tan(x+h)-tanx)/h = lim_(h->0) 1/h((tanx+tanh)/(1-tanxtanh)-tanx)#

#lim_(h->0) (tan(x+h)-tanx)/h = lim_(h->0) 1/h((canceltanx+tanh-canceltanx+tan^2xtanh)/(1-tanxtanh))#

#lim_(h->0) (tan(x+h)-tanx)/h = lim_(h->0) tanh/h((1+tan^2x)/(1-tanxtanh))#

#lim_(h->0) (tan(x+h)-tanx)/h = lim_(h->0) (sinh/h)(1/cosh)((1+tan^2x)/(1-tanxtanh))#

#lim_(h->0) (tan(x+h)-tanx)/h = 1*1*((1+tan^2x)/(1-tanx*0)) =1+tan^2x#

As:

#1+tan^2x = 1+sin^2x/cos^2x = (cos^2x+sin^2x)/cos^2x = 1/cos^2x#

putting it together we have:

#d/dx(x^2-tanx) = 2x-1/cos^2x#