Differentiate $f(x) = x^4 \tan^{-1}(6x)$. (Note: You can enter "arctan" or "tan$^($-1)" for the inverse tangent.)

Neetesh Kumar | November 08, 2024

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This is the solution to DHW Calculus
Assignment: 1 Question Number 17
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Step-by-step solution:

To find $f'(x)$, we’ll use the product rule for differentiation, as the function is a product of $x^4$ and $\tan^{-1}(6x)$.

The product rule states that if $f(x) = u(x) \cdot v(x)$, then $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$.

In this case:

• $u(x) = x^4$ and $v(x) = \tan^{-1}(6x)$.

Step 1: Differentiate $u(x)$ and $v(x)$

1. Differentiate $u(x) = x^4$: $u'(x) = 4x^3$

2. Differentiate $v(x) = \tan^{-1}(6x)$:

   • Use the chain rule here, where the derivative of $\tan^{-1}(x)$ is $\frac{1}{1 + x^2}$. $v'(x) = \frac{1}{1 + (6x)^2} \cdot 6 = \frac{6}{1 + 36x^2}$

Step 2: Apply the Product Rule

Using the product rule: $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

Substitute $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$: $f'(x) = 4x^3 \cdot \tan^{-1}(6x) + x^4 \cdot \frac{6}{1 + 36x^2}$

Final Answer:

$f'(x) = 4x^3 \tan^{-1}(6x) + \frac{6x^4}{1 + 36x^2}$

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