3 Differentiate the following with respect to x y = π ln x + 2.3⁵ˣ + 4/cotx + 2log₃x + 1/4x + 1/x⁶

The given expression is $\frac{1}{(4x+6)^{-6}}$.
This can be rewritten as $(4x+6)^6$.
Here, $a = 4x$, $b = 6$, and $n = 6$.
The binomial theorem states that $(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots$

The expansion to three terms is:

• For $\pi \ln x$: $\frac{d}{dx}(\pi \ln x) = \pi \cdot \frac{1}{x} = \frac{\pi}{x}$
• For $2 \cdot 3^{5x}$: Use the rule $\frac{d}{dx}(a^{kx}) = k \cdot a^{kx} \ln a$. $\frac{d}{dx}(2 \cdot 3^{5x}) = 2 \cdot (5 \cdot 3^{5x} \ln 3) = 10 \cdot 3^{5x} \ln 3$
• For $4 \tan x$: $\frac{d}{dx}(4 \tan x) = 4 \sec^2 x$
• For $\frac{2 \ln x}{\ln 3}$: $\frac{d}{dx}\left(\frac{2}{\ln 3} \ln x\right) = \frac{2}{\ln 3} \cdot \frac{1}{x} = \frac{2}{x \ln 3}$
• For $\frac{1}{4}x^{-1}$: $\frac{d}{dx}\left(\frac{1}{4}x^{-1}\right) = \frac{1}{4}(-1)x^{-2} = -\frac{1}{4x^2}$
• For $x^{-6}$: $\frac{d}{dx}(x^{-6}) = -6x^{-7} = -\frac{6}{x^7}$

The final answer is:

Let $u = 4 \cos x$ and $v = 10^{2x}$.

$u' = \frac{d}{dx}(4 \cos x) = -4 \sin x$
$v' = \frac{d}{dx}(10^{2x})$. Using the rule $\frac{d}{dx}(a^{kx}) = k \cdot a^{kx} \ln a$, we get:
$v' = 2 \cdot 10^{2x} \ln 10$

Factor out $4 \cdot 10^{2x}$ from the numerator:

Cancel $10^{2x}$ from the numerator and denominator:

The final answer is: