Find the tangent line to f(x)=7x+4ex at x=0

Question

Find the tngent line to f(x)=3x^2ln x at x=1

Find the tangent line to f(x) = ln(x) log2(x) at x=2

Answer 1

Well, I will start one of them

tangent line to f(x)=7x+4e^x at x=0 I assume you mean

slope = dy/dx = m = 7 + 4 e^x
when x = 0
slope = m = 7 + 4 = 11
so
y = 11 x + b
when x = 0, y = 4
so
4 = 11* (0) + b
so b = 4
y = 11 x + 4

Answer 2

1.
I will interpret that as f(x)=7x+4e^x at x=0
f'(x) = 7 + 4e^x
f'(0) = 7 + 4 = 11
f(0) = 0+4 = 4

equation of tangent:
(y-4) = 11(x-0)
arrange to whatever form you need

2.
f(x)= (3x^2)(ln x) at x=1
use product rule and follow steps from above

3.
I will interpret your log2(x) as log₂ x
recall that log₂ x
= lnx / ln2 or (1/ln2) lnx
so now you have another product rule like in #2

Answer 3

#3. No product rule needed, since ln2 is just a constant.