Question

The point in the interval $[0,2\pi ]$, where $f(x) = {e^x}\sin (x)$ has a maximum slope
$
  1)\,\pi /4 \\
  2)\,\pi /2 \\
  3)\,\pi \\
  4)\,3\pi /2 \\
 $

Answer 1

Hint: This a question based on basic concepts of differentiation. Firstly, the formula for the slope for any curve $y = f(x)$ is $\dfrac{{dy}}{{dx}}$. Secondly, to find the maximum value of the function we need to find the critical points using differentiation and lastly double differentiate the final equation to confirm whether the point will give a maxima or a minima.

Complete step-by-step solution:
Now we have, $f(x) = {e^x}\sin (x)$
 So, slope $ = $$f'(x)$
$ = $ $\dfrac{{d({e^x}\sin (x))}}{{dx}}$
Now, we use the product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$ to find the derivative of expression. So, we get,
$ = $ ${e^x}\dfrac{{d\left( {\sin x} \right)}}{{dx}} + \sin x\dfrac{{d\left( {{e^x}} \right)}}{{dx}}$
Now, we know that the derivative of $\sin x$ with respect to x is $\cos x$. Also, the derivative of ${e^x}$ with respect to x is ${e^x}$. So, we have,
$ = $ ${e^x}\cos x + {e^x}\sin x$
$ = $${e^x}(\sin (x) + \cos (x))$
So, the slope of the curve is represented by the expression ${e^x}(\sin (x) + \cos (x))$. Now, we need to find the critical points for the slope
So, $f''(x)$$ = $ $0$
$ \Rightarrow \dfrac{{d({e^x}(\sin (x) + \cos (x))}}{{dx}} = 0$
Opening the brackets,
$ \Rightarrow \dfrac{{d({e^x}\sin x + {e^x}\cos x)}}{{dx}} = 0$
$ \Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\sin x} \right) + \dfrac{d}{{dx}}\left( {{e^x}\cos x} \right) = 0$
Again using the product rule of differentiation, we get,
\[ \Rightarrow \left[ {{e^x}\dfrac{d}{{dx}}\left( {\sin x} \right) + \left( {\sin x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)} \right] + \left[ {{e^x}\dfrac{d}{{dx}}\left( {\cos x} \right) + \left( {\cos x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)} \right] = 0\]
Substituting the derivative of $\sin x$, $\cos x$ and ${e^x}$, we get,
\[ \Rightarrow \left[ {{e^x}\cos x + \left( {\sin x} \right)\left( {{e^x}} \right)} \right] + \left[ {{e^x}\left( { - \sin x} \right) + \left( {\cos x} \right)\left( {{e^x}} \right)} \right] = 0\]
\[ \Rightarrow {e^x}\cos x + {e^x}\sin x - {e^x}\sin x + {e^x}\cos x = 0\]
Simplifying the expression, we get,
$ \Rightarrow 2{e^x}\cos (x) = 0$
Dividing both sides of the equation by two, we get,
$ \Rightarrow {e^x}\cos (x) = 0$
Now, we know that exponential function cannot be equal to zero. So, we get, $\cos (x) = 0$.
Now, we know that cosine function is zero for odd multiples of $\left( {\dfrac{\pi }{2}} \right)$.
Hence, $x = \left( {2n + 1} \right)\dfrac{\pi }{2}$
Now, substituting in the values of n as zero and one.
$ \Rightarrow x = \left( {2 \times 0 + 1} \right)\dfrac{\pi }{2}$ and $x = \left( {2 \times 1 + 1} \right)\dfrac{\pi }{2}$
Doing the calculations, we get,
$ \Rightarrow x = \left( {2 \times 0 + 1} \right)\dfrac{\pi }{2}$ and $x = \left( {2 \times 1 + 1} \right)\dfrac{\pi }{2}$
$ \Rightarrow x = \dfrac{\pi }{2}$ and $x = \dfrac{{3\pi }}{2}$
Now, to check which critical point will the maxima or minima, we need to differentiate the above equation again
 So, $f'''(x) = \dfrac{{d(2{e^x}\cos (x))}}{{dx}}$
Using the product rule of differentiation, we get,
$ \Rightarrow f'''(x) = 2\left[ {{e^x}\dfrac{{d(\cos x)}}{{dx}} + \cos x\dfrac{{d\left( {{e^x}} \right)}}{{dx}}} \right]$
Substituting the values of derivatives,
$ \Rightarrow f'''(x) = 2\left[ { - {e^x}\sin x + {e^x}\cos x} \right]$
Taking common terms outside the bracket, we get,
$ \Rightarrow f'''(x) = 2{e^x}\left[ {\cos x - \sin x} \right]$
So, substituting both the values of x obtained as critical points, we get,
$f'''\left( {\dfrac{\pi }{2}} \right) = 2{e^{\left( {\dfrac{\pi }{2}} \right)}}\left[ {\cos \left( {\dfrac{\pi }{2}} \right) - \sin \left( {\dfrac{\pi }{2}} \right)} \right]$
We know the values of cosine and sine for $\dfrac{\pi }{2}$ radians.
$ \Rightarrow f'''\left( {\dfrac{\pi }{2}} \right) = 2{e^{\left( {\dfrac{\pi }{2}} \right)}}\left[ {0 - 1} \right]$
$ \Rightarrow f'''\left( {\dfrac{\pi }{2}} \right) = - 2{e^{\left( {\dfrac{\pi }{2}} \right)}}$
Similarly, $f'''\left( {\dfrac{{3\pi }}{2}} \right) = 2{e^{\left( {\dfrac{{3\pi }}{2}} \right)}}\left[ {\cos \left( {\dfrac{{3\pi }}{2}} \right) - \sin \left( {\dfrac{{3\pi }}{2}} \right)} \right]$
We know the values of cosine and sine for $\dfrac{{3\pi }}{2}$ radians.
$ \Rightarrow f'''\left( {\dfrac{{3\pi }}{2}} \right) = 2{e^{\left( {\dfrac{{3\pi }}{2}} \right)}}\left[ {0 - \left( { - 1} \right)} \right]$
$ \Rightarrow f'''\left( {\dfrac{{3\pi }}{2}} \right) = 2{e^{\left( {\dfrac{{3\pi }}{2}} \right)}}$
Now, we know that exponential function is always positive.
So, the function has maxima for $x = \dfrac{\pi }{2}$ as the second derivative is negative for the value. Similarly, the function has minima for $x = \dfrac{{3\pi }}{2}$ as the second derivative is positive for the value.
So, the maximum value of slope is for $x = \dfrac{\pi }{2}$.

Note: We need to know the product rule and basics of differentiation before solving such questions. We must know the expressions for the slope and normal of curves to get to the required answer. Care should be taken while handling the calculative steps. Simplification rules may be used to ease the calculations and to deal with expressions.