Can anybody help me with my calc. homework? I suck at math?

Question

1. Find an equation of the tangent line to the graph of f(x)=x ln(2x+3) at the point (-1,0)
2. Use logarithmic diffrentiation to compute y' , given y=(2x+1)^3(3x+4)^5

Answer 1

1.
-Use the Product Rule:

ƒ' (x) = d/dx [ x ln (2x +3) ]
ƒ' (x) = x d/dx ln (2x +3) + ln (2x + 3) * d/dx (x)
ƒ' (x) = x ( 2 / 2x + 3 ) + ln ( 2x + 3 ) * 1
ƒ' (x) = ( 2x / 2x + 3 ) + ln ( 2x + 3 )

-Slope of a tangent line to the graph of ƒ at (-1, 0):

ƒ' ( -1 ) = ( -2 / -2 + 3 ) + ln 1
ƒ' ( -1 ) = -2

-Then, find the equation:

y - 0 = -2 (x + 1)
y = -2x - 2


2.
-Logarithm on both sides:

ln y = ln (2x + 1)³ (3x + 4)^5
ln y = ln (2x + 1)³ + ln (3x + 4)^5
ln y = 3 ln (2x + 1) + 5 ln (3x+4)

-Differentiate with respect to x

d/dx (ln y) = y' / y = 3 * (2 / 2x + 1) + 5 * (3 / 3x + 4)
d/dx (ln y) = 3 [ (2 / 2x + 1) + (5 / 3x + 4) ]
...
y' = 3 (2x + 1)³ (3x+4)^5 [ (2 / 2x + 1) + (5 / 3x + 4) ]


P.S: are you going to be able to pass the class like this?

Answer 2

1. f'(x)= x (1/(2x+3))(2) + ln(2x+3)
f'(-1)= -1(1)(2)+ln( 1) = -2 slope of tangent line,
so
y=-2(x+1)

2. y=(2x+1)^3(3x+4)^5
ln y = ln [(2x + 1)³ (3x + 4)^5]
ln y = ln (2x + 1)³ + ln (3x + 4)^5
ln y = 3 ln (2x + 1) + 5 ln (3x+4)

take the derivative with respect to x:

y' / y = 3 (2 / 2x + 1) + 5 (3 / 3x + 4)
y' = (2x + 1)³ (3x+4)^5 [ 3(2 / 2x + 1) + 5(3 / 3x + 4) ]