can someone please help me with these 3 questions:
1) Evaluate the limit.
lim_(x->infinity) (ln(ln(4 x)))/(4 x)
2)Find y ' and y ".
y= ln(5 x)/x^7
3)y =(arctan)(x - sqrt(1+x^2))
for this question i got y = -1/(2(1+x^2))which was wrong.
Please and thank you!
Certainly! I can help you with these questions. Let's go through each question step by step and explain how to solve them.
1) Evaluate the limit:
To evaluate the limit, we can use L'Hôpital's rule. First, let's rewrite the expression as follows:
lim_(x->infinity) [(ln(ln(4 x)))/(4 x)]
Now, we can take the derivative of the numerator and the denominator with respect to x:
lim_(x->infinity) [(1/(ln(4 x)))*(1/(4 x))* ((4 x)/x)]
= lim_(x->infinity) [(1/4x) (1/ln(4 x))]
Again, we can take the derivative of the numerator and the denominator with respect to x:
= lim_(x->infinity) [-1/(16x^2 ln(4 x))]
Now, substitute x with infinity:
lim_(x->infinity) [-1/(16x^2 ln(4 x))] = 0
Therefore, the limit of (ln(ln(4 x)))/(4 x) as x approaches infinity is equal to 0.
2) Find y' and y":
To find the first derivative of y= ln(5 x)/x^7, we can use the quotient rule. The quotient rule states that if y = f(x)/g(x), then y' = (f'(x)g(x) - f(x)g'(x))/(g(x))^2
Let's apply the quotient rule to find y':
y = ln(5 x)/x^7
y' = [(1/x^7)(5 x) - ln(5 x)(7x^6)]/(x^7)^2
= [5/x^6 - 7ln(5 x)/x]/x^14
= (5 - 7ln(5 x))/x^7
Now, let's find y":
Using the quotient rule again, we can find y":
y' = (5 - 7ln(5 x))/x^7
y" = [(0 - 7/x)(x^7) - (5 - 7ln(5 x))(7x^6)]/(x^7)^2
= [-7x + 49x^6 - 49x^6ln(5 x)]/x^14
= (-7x - 49x^6(1 + ln(5 x)))/x^14
= -7/x - 49(1 + ln(5 x))/x^8
So, the first derivative of y is (5 - 7ln(5 x))/x^7 and the second derivative is -7/x - 49(1 + ln(5 x))/x^8.
3) Simplify y =(arctan)(x - sqrt(1+x^2)):
To simplify this expression, we need to use the trigonometric identity for arctan. The identity is arctan(u) = arctan(1/u) + π/2
Let's rewrite the expression:
y = arctan(x - sqrt(1+x^2))
= arctan[(1/(x + sqrt(1+x^2)))]
= arctan(1/[x + sqrt(1+x^2)]) + π/2
Therefore, the simplified expression of y is arctan(1/[x + sqrt(1+x^2)]) + π/2.
I hope this explanation helps! Let me know if you have any further questions.