Find the derivative of #f(x)=-tan(x)#?

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Answer 1 · 2018-02-12T14:46:52

#f'(x)=-sec^2(x)#

We want to find the derivative of

#f(x)=-tan(x)#

Use the definition #tan(x)=sin(x)/cos(x)#

#f(x)=-sin(x)/cos(x)#

Use the quotient rule, if #f(x)=(h(x))/g(x)#,

then #f'(x)=(h'(x)g(x)-h(x)g'(x))/(h(x))^2#

By the quotient rule with #h(x)=sin(x)# and #g(x)=cos(x)#

#f'(x)=-((d/dx(sin(x)))cos(x)-sin(x)(d/dx(cos(x))))/cos^2(x)#

#=-(cos(x)cos(x)+sin(x)sin(x))/cos^2(x)#

#=-(cos^2(x)+sin^2(x))/cos^2(x)#

#=-1/cos^2(x)#

#=-sec^2(x)#