Question

How do you find the equation of the tangent line to graph `y=x^sinx` at the point `(pi/2, pi/2)`?

Answer 1

The equation of the tangent line is `y= x`.

Explanation:

`lny = ln(x^sinx)`

`lny= sinxlnx`

Now, use the product rule and implicit differentiation to find the derivative.

`1/y(dy/dx) = (cosx xx lnx + sinx xx 1/x)`

`1/y(dy/dx) = cosxlnx + sinx/x`

`dy/dx = (cosxlnx + sinx/x)/(1/y)`

`dy/dx = y(cosxlnx + sinx/x)`

Since `y = x^sinx`:

`dy/dx = x^sinx(cosxlnx + sinx/x)`

All that is left to do is find the slope of the tangent. This can be found by inserting `x = a` into the derivative, since the derivative represents the instantaneous rate of change of the function.

`m_"tangent" = (pi/2)^(sin(pi/2))(cos(pi/2)ln(pi/2) + (sin(pi/2))/(pi/2))`

`m_"tangent" = pi/2(0 + 1/(pi/2))`

`m_"tangent" = pi/2(0 + 2/pi)`

`m_"tangent" = 1`

Now, we use point-slope form to find the equation of the line.

`y - y_1 = m(x - x_1)`

`y- pi/2 = 1(x- pi/2)`

`y - pi/2 = x - pi/2`

`y = x - pi/2 + pi/2`

`y = x`

Hopefully this helps!