How to find the gradient of the tangent to this curve?

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Answer 1 · 2018-03-18T13:56:20

Gradient at #(0,1)#

#color(blue)(9/2)#

Gradient at #(1,e^(1/2)+4)#

#color(blue)(1/2e^(1/2)+4~~4.8244)#

#f(x)=e^(x/2)+4x#

Rewrite:

#e^(x/2)=(e^x)^(1/2)#

So we have:

#f(x)=(e^x)^(1/2)+4x#

We need to differentiate this to get the gradient function.

We need to use the chain rule for #(e^x)^(1/2)#

Let #u=e^x#

Then by the chain rule:

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

#dy/dx((e^x)^(1/2)+4x)=1/2u^(-1/2)*u+4#

Substituting #u=e^x#

#1/2e^(-x/2)*e^x+4=1/2e^(x/2)+4#

Gradient at #(0,1)#

#1/2e^(0/2)+4=color(blue)(9/2)#

Gradient at #(1,e^(1/2)+4)#

#color(blue)(1/2e^(1/2)+4~~4.8244)#