Find the equation of the tangent to the

gabolzm6d

gabolzm6d

Answered question

2022-04-12

Find the equation of the tangent to the curve y=2x+2-x at the point (2,414)
y=2x+2-x
lny=xln2-xln2
lny=0
1ydydx=0
dydx=0
I've checked the answer and I've got the differential wrong, What am I doing wrong? I assume it has something to do with the expression with the negative exponent? Am I not allowed to prefix the natural expression with "minus" x? Could someone explain please, thank you!

Answer & Explanation

Dubliddissefyf8

Dubliddissefyf8

Beginner2022-04-13Added 10 answers

Your second step is incorrect:
ln(y)=ln(2x+2x)
but
ln(y)ln(2x)+ln(2x)
You don't need logs here, just differentiate each side as normal, using the following on the RHS:
ddx(f(x)+g(x))=ddxf(x)+ddxg(x)
Buizzae77t

Buizzae77t

Beginner2022-04-14Added 13 answers

Again, no need to take logs.
y=ln2(2x+2x).For x=2 this gives y=ln2174 .
This is your slope. Use this to solve for the intercept n=174(ln2174)2

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