Calculus Tangent Line

[lala_doda]

Can someone please explain step-by-step how to do these?

Give the equation of the tangent line to the given graph at the point where x=0

a. f(x)=ln(6x+1)+ e^(2x)
b. f(x)=ln(2x+1)- 3e^(-4x)

the answers are:
a. y-1=8(x-0)
b. y+3=14(x-0)

 

[pka]

Can someone please explain step-by-step how to do these?
Give the equation of the tangent line to the given graph at the point where x=0

The equation in each case is:
\(\displaystyle (y-f(0))=f'(0)x.\)

 

[burakaltr]

Can someone please explain step-by-step how to do these?

Give the equation of the tangent line to the given graph at the point where x=0

a. f(x)=ln(6x+1)+ e^(2x)
b. f(x)=ln(2x+1)- 3e^(-4x)

the answers are:
a. y-1=8(x-0)
b. y+3=14(x-0)

Well, you need to find the derivatives evaluated at x = 0

Then use the fact that y-y1=(derivativeat0)(x-x1)

Hint : x1 and y1 are the values where x=0 and y are the straightforwardly the original functions themselves evaluated at x=0

Good Luck !

 

[lala_doda]

got it. thank you!