Answer
y = $2sin(2)x$ - $2sin(2)$
Work Step by Step
To find the equation of a tangent line to the graph f(x) at the given point, we must first take the derivative of f(x).
(a) f(x) = $sin(2x)\ln(x^{2})$
Using the product rule and chain rule, we get:
f'(x) = $sin(2x)\times\frac{1}{x^{2}}\times2x + cos(2x)\times 2\times \ln(x^{2})$
Multiplying out, out final derivative is:
f'(x) = $\frac{2sin(2x)}{x} + 2cos(2x)\ln(x^{2})$
Plugging in the point's x-value of $1$, we get:
f'(x) = $\frac{2sin(2)}{1} + 2cos(2)\ln(1^{2})$ or f'(x) = $2sin(2)$
Therefore, we are able to write the tangent line in point-slope form as:
(y-0) = $2sin(2)$(x-1)
Simplifying, we get:
y = $2sin(2)x$ - $2sin(2)$
To graph the function and its tangent line at the point, use Desmos.
(b) See graph of the original function f(x) (the oscillating function), its tangent line f'(x) (the straight line, and the point at which f'(x) touches f(x).
To confirm results, use any graphing software's derivative function.
(c) Methods and results vary.
