Question:

Let $f(x)$ be a function defined by $f(x) = 3x^4 - 2\ln(x) + 5\sin(x)$.

(a) Find $f'(x)$.

(b) Find the equation of the tangent line to the graph of $f(x)$ at the point where $x = \pi/2$.

Answer:

(a) To find the derivative of $f(x)$, we will need to apply the rules of differentiation for each term in the function.

The derivative of a constant is always zero, so we can disregard the constant term and focus on differentiating the other terms of $f(x)$.

The power rule states that if we have a term of the form $x^n$ where $n$ is a constant, the derivative is given by $nx^{n-1}$. Applying the power rule, we find:

The derivative of the natural logarithm function $\ln(x)$ is $\frac{1}{x}$. So, differentiating $-2\ln(x)$, we get:

Finally, the derivative of the sine function $\sin(x)$ is the cosine function $\cos(x)$. Thus, differentiating $5\sin(x)$, we have:

Adding all the derivatives together, we get:

Therefore, the derivative of $f(x)$ is given by $f'(x) = 12x^3 - \frac{2}{x} + 5\cos(x)$.

(b) To find the equation of the tangent line to the graph of $f(x)$ at the point $x = \frac{\pi}{2}$, we need two pieces of information: the slope of the tangent line and a point it passes.

The slope of the tangent line can be determined by evaluating $f'(x)$ at $x = \frac{\pi}{2}$. Let's find $f'\left(\frac{\pi}{2}\right)$:

Simplifying this expression gives:

So, the slope of the tangent line is $\frac{3\pi^2}{2} - \frac{4}{\pi}$.

Now, we need to find a point on the graph of $f(x)$ that the tangent line passes through. This point will be $P\left(\frac{\pi}{2}, f\left(\frac{\pi}{2}\right)\right)$. Evaluating $f\left(\frac{\pi}{2}\right)$, we get:

Simplifying this expression gives:

Finally, we can write the equation of the tangent line using the slope-point form: $y - y_1 = m(x - x_1)$. Plugging in the values we found:

Hence, the equation of the tangent line to the graph of $f(x)$ at the point $x = \frac{\pi}{2}$ is: