Differentiate with respect to $$ x $$ if $$ y=2 \log x- an ^{2} x-3 r^{2 x}+x^{-2}-3^{-2 x}+9 x+6 $$

Question

Differentiate with respect to $$ x $$ if $$ y=2 \log x-	an ^{2} x-3 r^{2 x}+x^{-2}-3^{-2 x}+9 x+6 $$

UpStudy AI · Accepted Answer

$$ y = 2\ln x - \tan ^2 x - 3r^{2x} + x^{-2} - 3^{-2x} + 9x + 6 $$

We differentiate term by term.

1. For $$ 2\ln x $$:
   $$
   \frac{d}{dx}(2\ln x) = \frac{2}{x}.
   $$

2. For $$-\tan^2 x$$:
   $$
   \frac{d}{dx}(\tan^2 x) = 2\tan x \sec^2 x,
   $$
   so,
   $$
   \frac{d}{dx}(-\tan^2 x) = -2\tan x \sec^2 x.
   $$

3. For $$-3r^{2x}$$:  
   Treat $$ r $$ as a constant. Recall the derivative of $$ r^{2x} $$ is:
   $$
   \frac{d}{dx}(r^{2x}) = r^{2x}\ln (r) \cdot 2.
   $$
   Then,
   $$
   \frac{d}{dx}(-3r^{2x}) = -3\cdot 2\ln (r) \, r^{2x} = -6\ln (r)\, r^{2x}.
   $$

4. For $$ x^{-2} $$:
   $$
   \frac{d}{dx}(x^{-2}) = -2x^{-3}.
   $$

5. For $$-3^{-2x}$$:  
   Write $$ 3^{-2x} $$ as an exponential. The derivative of $$ 3^{-2x} $$ is:
   $$
   \frac{d}{dx}(3^{-2x}) = 3^{-2x}\ln (3)\cdot (-2) = -2\ln (3) \, 3^{-2x}.
   $$
   Hence,
   $$
   \frac{d}{dx}(-3^{-2x}) = 2\ln (3) \, 3^{-2x}.
   $$

6. For $$ 9x $$:
   $$
   \frac{d}{dx}(9x) = 9.
   $$

7. For the constant $$ 6 $$:
   $$
   \frac{d}{dx}(6) = 0.
   $$

Combining all these results, the derivative is:
$$
y' = \frac{2}{x} - 2\tan x\,\sec^2 x - 6\ln(r)\,r^{2x} - 2x^{-3} + 2\ln(3)\,3^{-2x} + 9.
$$
The derivative of $$ y $$ with respect to $$ x $$ is: $$ y' = \frac{2}{x} - 2\tan x\,\sec^2 x - 6\ln(r)\,r^{2x} - \frac{2}{x^3} + 2\ln(3)\,3^{-2x} + 9 $$

Differentiate with respect to \( x \) if \( y=2 \log x-\tan ^{2} x-3 r^{2 x}+x^{-2}-3^{-2 x}+9 x+6 \)

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