How do you find the tangent line to #x=cos(y/4)# at #y=π#?

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Answer 1 · 2018-04-07T00:50:38

#-4sqrt2=dy/dx#

We have:

#x=cos(y/4)# Let's let #f(x)=y#.

#=>x=cos(f(x)/4)# Apply derivative on both sides.

#=>d/dx(x)=d/dx(cos(f(x)/4))#

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.

#d/dx(f(g(x)))=f'(g(x))*g'(x)#

#d/dx(cos(x))=-sin(x)#

#=>1*x^(1-1)=-sin(f(x)/4)*d/dx(f(x)/4)#

#=>x^(0)=-sin(f(x)/4)*1/4*d/dx(f(x))#

#=>4=-sin(f(x)/4)*f'(x)#

#=>4/-sin(f(x)/4)=f'(x)# Remember that #f(x)=y# and #f'(x)=dy/dx#

#=>4/-sin(y/4)=dy/dx#

When #y=pi...#

#=>4/-sin(pi/4)=dy/dx#

#=>4/(-sqrt2/2)=dy/dx#

#=>4*-2/sqrt2=dy/dx#

#=>-8/sqrt2=dy/dx#

#=>-4sqrt2=dy/dx#

That is the answer!