Differential Calculus using log and ln

Question 1

y = ln(3x)?

Answer 1

rule: y = lnx | dy/dx = 1/x

3/3x

cross out

1/x.

Question 2

What is the derivative of
y = lnx^3?

Answer 2

3/x

Question 3

y = ln^3x

Answer 3

pay attention

let u = lnx

let y = u^3

u’= 1/x

y’ = 3u^2

chain rule

3(lnx)^2/x

that it.

Question 4

ln^2(12x^2+1)

Answer 4

Simplify and break it down.

y = ln^2(12x^2 + 1) –> (ln(12^2+1))^2

let m = ln(12x^2+1)

so y = m^2

now, within m it can be further simplified since we have 12x^2+1

so let n = 12x^2+1

Now we have

y = m^2 –> 2m

m = ln(n) –> 1/n

n = 12x^2 +1 –> 24x

times them all together, thus chain rule

48x(ln12x^2+1)/12x^2+1

48x becomes we did the chain rule, 24 times 2

Question 5

y = log(x-5/2-7x)

Answer 5

bring it out

y = log(x-5) - log(2-7x)

so lets do log(x-5)

break it up

let u = x-5 –> 1

so log(u)

log(u) is 1/u

u is?

x-5

so 1/x-5 times 1 = 1/x-5

However because of the log, you also must add ln 10 because by default, log has a base 10

Any base must be in the denominator as part of the rule

y = log_a x | y= 1/xlna

so

1/x-5ln10

now log(2-7x)

this becomes

so let n = 2-7x –. 7

y = log(n) – > 1/n

chain rule

7/2-7xln10

overall

1/x-5ln10 - 7/2-7xln10

Question 6

y = 5^x/log_5(x^2+9)

Answer 6

BREAK IT DOWN.

See a fraction? QUOTIENT RULE

quot rule: y = u’v - v’u/v^2

so u = 5^x

v = log_5(x^2+9)

derive u = 5^x ln5 due to rule y = a^x | y’ = a^x lna

derive v

let u = x^2+9 –> 2x

v = log_5(u) = 1/uln5

times

v = 2x/x^2+9ln5

now sub in quot rule

Question 7

y = e^5x

Answer 7

break it up

y= e^u – stays the same

u = 5x –> 5

chain rule

5e^5x

Question 8

y = sqrt(e^7x^5)

Answer 8

remove the sqrt

y = (e^7x^5)^1/2

now break it down, is a chain rule so

y = u^1/2 – > 1/2 m ^ -1/2, because of negative exponent, it can become 1/2sqrt(u)

u = e^7x^5

too much

so u = e^n – > n e

m = 7x^5 –> 35x^4

times it all together and sub it in

Question 9

y = e^sqrt(x)

Answer 9

y = e^n –> stays the same since eular number

n= sqrt(x) –> 1/2 x^-1/2 becomes 1/2sqrt(x)

chain rule

e^sqrt(x) / 2sqrt(x)

Question 10

find the equation of tangent and normal lines to the curve

y = ln(2x-1) when x = 4/

Answer 10

so y = mx+b

y = ln(2(4)-1) = ln(7)

m = ln(2x-1)

let u = 2x-1

therefore m’ = ln(u) = 1/u

u = 2x-1 = 2

chain rule

2/2x-1

sub x.

2/7

therefore

ln(7) = 2/7(4) + b

find b

ln(7) - 8/7 ( times 4) = b

so overall tangent equation is

y = 2x/7 + (ln(7) - 8/7

normal line =

m_t times m_n = -1

m_t = 2/7

so

2/7 times m_n = -1

algebra

m_n = -7/2

ln(7) = -7/2(4) + b

therefore

y = -7x/2 + ln(7) + 14

SO

tangent line:

y = 2/7 + ln(7) - 8/7

normal line:

y = -7x/2 + ln(7) + 14