1.) Find the derivative of ((x + 3)/e^x) ^ (Log e X)

2.) T/F (log e X)' = (log e |x|)'

3.) T/F (Log e 7x)' = (log e x)'

4.) Find the derivative of e^(log e 7x)

y = ((x+3)/e^x)^lnx

wow!
lny = lnx(ln(x+3)-ln(e^x))
= lnx(ln(x+3)-1)
1/y y' = 1/x (ln(x+3)-1) + lnx*1/(x+3)
y' = y(ln(x+3)/x - 1/x + lnx/(x+3))
Now you can massage that in many ways.

#2 since lnx is not defined for x<0, I'd say False

#3 since ln 7x = ln7 + lnx, I'd say True

#4 7 (why?)

1.) To find the derivative of the given expression, we can use the chain rule. The chain rule states that if we have a composition of functions, such as (f(g(x))), the derivative is given by (f'(g(x)) * g'(x)).

Let's break down the expression ((x + 3)/e^x)^(LogeX):

Step 1: Rewrite the expression using exponentiation notation:
y = ((x + 3)/e^x)^(LogeX)
y = e^(LogeX * Log((x + 3)/e^x))

Step 2: Take the natural logarithm of both sides to simplify:
ln(y) = LogeX * Log((x + 3)/e^x)

Step 3: Differentiate implicitly with respect to x:
(1/y) * (dy/dx) = (d/dx)[LogeX * Log((x + 3)/e^x)]

Step 4: Apply the chain rule to the right-hand side:
(1/y) * (dy/dx) = ((d/dx)[LogeX] * Log((x + 3)/e^x)) + (LogeX * (d/dx)[Log((x + 3)/e^x)])

Step 5: Find the derivatives of LogeX and Log((x + 3)/e^x):
(d/dx)[LogeX] = 1/X (using the derivative of natural logarithm)
(d/dx)[Log((x + 3)/e^x)] = (1/((x + 3)/e^x)) * ((d/dx)[(x + 3)/e^x]) (using the chain rule)
= (1/((x + 3)/e^x)) * (((e^x * (d/dx)[(x + 3)]) - ((x + 3) * (d/dx)[e^x]))/e^(2x))
= ((e^x * 1) - ((x + 3) * (-e^x)))/(x + 3)
= (e^x + (x + 3)e^x)/(x + 3)
= (e^x(1 + x + 3))/(x + 3)

Step 6: Substitute back into the equation and solve for (dy/dx):
(1/y) * (dy/dx) = (1/X * Log((x + 3)/e^x)) + (LogeX * (e^x(1 + x + 3))/(x + 3))

Step 7: Solve for (dy/dx) by multiplying both sides by y:
dy/dx = y * [(1/X * Log((x + 3)/e^x)) + (LogeX * (e^x(1 + x + 3))/(x + 3))]

Step 8: Substitute back the original expression for y and simplify:
dy/dx = ((x + 3)/e^x)^(LogeX) * [(1/X * Log((x + 3)/e^x)) + (LogeX * (e^x(1 + x + 3))/(x + 3))]

2.) T/F (log e X)' = (log e |x|)'
True.

The derivative of the natural logarithm function, ln(x), is given by (1/x). Therefore, the derivative of ln(x) is (1/x), which is independent of the sign of x. So, whether we take the natural logarithm of x or |x|, the derivative remains the same.

3.) T/F (Log e 7x)' = (log e x)'
False.

The derivative of ln(7x) is not the same as the derivative of ln(x). To find the derivative of ln(7x) with respect to x, we need to use the chain rule. The chain rule states that the derivative of ln(u) is (1/u) * du/dx.

So, in this case, (Loge(7x))' = (1/(7x)) * (7) = 1/x.

Therefore, (Loge(7x))' ≠ (log e x)'.

4.) To find the derivative of e^(ln(7x)), we can use the chain rule.

Let's break it down step by step:

Step 1: Rewrite the expression as e^(ln(7x)).
Step 2: Apply the chain rule. The derivative of e^(ln(7x)) is equal to e^(ln(7x)) multiplied by the derivative of ln(7x).
Step 3: The derivative of ln(7x) can be calculated using the chain rule, which states that the derivative of ln(u) is (1/u) multiplied by the derivative of u.
Step 4: The derivative of 7x with respect to x is 7.

Putting it all together, the derivative of e^(ln(7x)) is e^(ln(7x)) * (1/(7x)) * 7.

Simplifying further, we get the derivative as e^(ln(7x)) * (1/x).

Since e raised to the power of ln(7x) is simply 7x, the derivative becomes 7x * (1/x), which simplifies to 7.