How do you find the derivative of #f(x)=ax+b#?

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Answer 1 · 2017-09-25T03:35:32

#f^'(x)=a#

#f(x)=ax+b#

Take derivative on both sides:

#d/dx(f(x))=d/dx(ax+b)#

Apply the sum/difference rule for derivative which is stated as:

#d/dx(f+g)=d/dx(f)+d/dx(g)#

So that we will have:

#f^'(x)=d/dx(ax)+d/dx(b)#

Remember the derivative of a constant is zero, so that we will have:

#f^'(x)=d/dx(ax)+0#

Take the constant out by applying #d/dx(a*f)=a*d/dx(f)#

#f^'(x)=a*d/dx(x)#

Apply the common derivative rule #d/dx(x)=1#

#f^'(x)=a*1#

#f^'(x)=a#