Q.Find $\frac{dy}{dx}$ in the following: $(5x)^{3 \cos 2x}$

Concept: Implicit Differentiation (with logarithmic differentiation, since the variable appears in both the base and the exponent).

Let $y = (5x)^{3 \cos 2x}$. Take the natural logarithm of both sides:

Differentiate both sides with respect to $x$ (using the product rule on the right):

Simplify the bracket:

Multiply through by $y = (5x)^{3 \cos 2x}$:

When you see a function where both the base and the exponent contain the variable — like $(5x)^{3\cos 2x}$ — the standard differentiation rules (power rule, exponential rule) don't apply directly. The power rule assumes a constant exponent; the exponential rule assumes a constant base. Here, both are moving.

The trick is to use logarithmic differentiation. By taking the natural log, we turn the exponent into a product, which we can then differentiate using the product rule and chain rule. This is the cleanest, most reliable method for this type of problem.

Let’s work through it.

1. Set up the equation. Let $y = (5x)^{3\cos 2x}$. Take the natural logarithm of both sides:

Using the power property of logs, $\log(a^b) = b \log a$, we get:

2. Differentiate both sides with respect to $x$. On the left, $\frac{d}{dx}[\log y] = \frac{1}{y} \cdot \frac{dy}{dx}$ (chain rule). On the right, we have a product: $3\cos 2x$ times $\log(5x)$. Use the product rule:

3. Compute the derivatives in the product.

   • For $\frac{d}{dx}[3\cos 2x]$: The derivative of $\cos 2x$ is $-\sin 2x \cdot 2 = -2\sin 2x$, so multiplied by 3 gives $-6\sin 2x$.
   • For $\frac{d}{dx}[\log(5x)]$: $\log(5x) = \log 5 + \log x$, so its derivative is $\frac{1}{x}$. (Or directly: derivative of $\log(5x)$ is $\frac{1}{5x} \cdot 5 = \frac{1}{x}$.)

   So the right-hand side becomes:

4. Put it together. We have:

5. Solve for $\frac{dy}{dx}$. Multiply both sides by $y$:

Now substitute back $y = (5x)^{3\cos 2x}$:

Method: Logarithmic Differentiation for $y=[f(x)]^{g(x)}$

When BOTH the base and the exponent of a power contain $x$, neither the power rule (needs constant exponent) nor the exponential rule (needs constant base) applies directly. Logarithmic differentiation converts the exponent into a product, which can then be handled with the ordinary rules.

Steps

Step 1: Take the natural log of both sides

using the power property of logs, $\log(a^b)=b\log a$.

Step 2: Differentiate both sides with respect to $x$

On the left, $\frac{d}{dx}\log y=\frac{1}{y}\frac{dy}{dx}$ (chain rule, since $y$ is a function of $x$). On the right, apply the product rule (since it's now $g(x)$ times $\log f(x)$), with the chain rule on $\log f(x)$ itself.

Step 3: Solve for $\dfrac{dy}{dx}$ by multiplying both sides by $y$

Step 4: Substitute the original expression for $y$ back in

Applying to this problem: for $y=(5x)^{3\cos2x}$, $\log y=3\cos2x\cdot\log(5x)$; differentiating the right side needs the product rule (with the chain rule bringing down a factor of $2$ from $\cos2x$'s argument, and $\log(5x)$ differentiating to $\frac{1}{x}$), giving $\frac{dy}{dx}=(5x)^{3\cos2x}\left(\frac{3\cos2x}{x}-6\sin2x\log(5x)\right)$.

Common Mistakes

Mistake 1: Trying to apply the power rule directly since the exponent "looks constant-ish".

Why it's wrong: $3\cos2x$ genuinely depends on $x$ — the power rule's requirement of a fixed exponent is not met, and applying it anyway gives a completely wrong derivative shape. Correct approach: always check whether the exponent contains $x$ before choosing a differentiation method.

Mistake 2: Forgetting the chain-rule factor of $2$ when differentiating $\cos2x$ inside the product-rule expansion.

Why it's wrong: $\frac{d}{dx}(3\cos2x)=-6\sin2x$, not $-3\sin2x$ — the inner $2x$ contributes its own factor of $2$. Correct approach: differentiate $\cos2x$ as its own mini chain-rule step before multiplying by the constant $3$.