Question

If $y = \sin x + {e^x}$, then $\dfrac{{{d^2}x}}{{d{y^2}}}$
A. $\dfrac{{\left( {\sin x - {e^x}} \right)}}{{{{\left( {\cos x + {e^x}} \right)}^3}}}$
B. ${\left( { - \sin x + {e^x}} \right)^{ - 1}}$
C. $\dfrac{{\left( {\sin x - {e^x}} \right)}}{{{{\left( {\cos x + {e^x}} \right)}^2}}}$
D. None of these

Answer 1

Here, the given question. We have to find the derivative or differentiated term of the function. For this, first consider function $y$, then differentiate $y$ with respect to $x$ and again or second time differentiate the function with respect to $y$ term by using a standard differentiation formula of trigonometric and exponential function and use chain rule for differentiation. And on further simplification we get the required differentiation value.
Formula used:
Here in this question, we used some standard differentiation formula i.e.,
$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
$\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$
$\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}$

Complete step by step answer:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
The Chain Rule is a formula for computing the derivative of the composition of two or more functions.
The chain rule expressed as $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}$
Consider the given function $y$
$\Rightarrow \,\,\,\,y = \sin x + {e^x}$---------- (1)
Differentiate function $y$ with respect to $x$
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\sin x + {e^x}} \right)$
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sin x} \right) + \dfrac{d}{{dx}}\left( {{e^x}} \right)$
On differentiating using a formula $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$ and $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$, then we have
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \cos x + {e^x}$ -----(2)
Now, to find the $\dfrac{{{d^2}x}}{{d{y^2}}}$. Express equation in terms of $\dfrac{{dx}}{{dy}}$ by taking reciprocal.
On taking reciprocal, then equation (2) becomes
$\Rightarrow \,\,\,\,\dfrac{{dx}}{{dy}} = \dfrac{1}{{\cos x + {e^x}}}$ ----(3)
Again, differentiate with respect to $y$
$\Rightarrow \,\,\,\,\dfrac{{dx}}{{dy}} = \dfrac{1}{{\cos x + {e^x}}}$
$\Rightarrow \,\,\,\,\dfrac{d}{{dy}}\left( {\dfrac{{dx}}{{dy}}} \right) = \dfrac{d}{{dy}}\left( {\dfrac{1}{{\cos x + {e^x}}}} \right)$
On differentiating using a formula $\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}$, and then by chain rule we have
$\Rightarrow \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{ - 1}}{{{{\left( {\cos x + {e^x}} \right)}^2}}}\dfrac{d}{{dy}}\left( {\cos x + {e^x}} \right)$
$\Rightarrow \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{ - 1}}{{{{\left( {\cos x + {e^x}} \right)}^2}}}\left[ {\dfrac{d}{{dy}}\left( {\cos x} \right) + \dfrac{d}{{dy}}\left( {{e^x}} \right)} \right]$
We know that $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$ and $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$, on applying we have
$\Rightarrow \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{ - 1}}{{{{\left( {\cos x + {e^x}} \right)}^2}}}\left[ { - \sin x\dfrac{{dx}}{{dy}} + {e^x}\dfrac{{dx}}{{dy}}} \right]$
$\Rightarrow \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{ - 1}}{{{{\left( {\cos x + {e^x}} \right)}^2}}}\left( { - \sin x + {e^x}} \right)\dfrac{{dx}}{{dy}}$
$\Rightarrow \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{ - \left( { - \sin x + {e^x}} \right)}}{{{{\left( {\cos x + {e^x}} \right)}^2}}} \cdot \dfrac{{dx}}{{dy}}$
Substitute $\dfrac{{dx}}{{dy}}$ by using equation (3), then
$\Rightarrow \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{ - \left( { - \sin x + {e^x}} \right)}}{{{{\left( {\cos x + {e^x}} \right)}^2}}} \times \dfrac{1}{{\cos x + {e^x}}}$
On simplification, we get
$\therefore \,\,\,\,\dfrac{{{d^2}x}}{{d{y^2}}} = \dfrac{{\left( {\sin x - {e^x}} \right)}}{{{{\left( {\cos x + {e^x}} \right)}^3}}}$
Hence, it’s a required solution.

So, the correct answer is “Option A”.

Note: The student must know about the differentiation formulas for the trigonometric, exponential, algebraic, logarithm functions and these differentiation formulas are standard. If the function is complex, we have to use the chain rule differentiation. It makes it easy to find out the differentiated term.