Question:

The slopes of the tangent and normal at $(0, 1)$ for the curve $y = \sin x + e^x$ are respectively

Updated On: Sep 12, 2023
  • $1$ and $-1$
  • $- \frac{1}{2}$ and $2$
  • $2$ and $- \frac{1}{2}$
  • $-1$ and $1^-$
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The Correct Option is C

Solution and Explanation

We have, $y = \sin x + e^x$
Differentiating w.r.t. x, we get
$\frac{dy}{dx} = \cos x +e^{x}$
$ \left|\frac{dy}{dx}\right|_{\left(0,1\right)} = \cos\left(0\right)+e^{0} =1+1=2$
$ - \frac{dx}{dy} = -\frac{1}{\cos x +e^{x}} $
$\left|-\frac{dx}{dy}\right|_{\left(0,1\right)} = -\frac{1}{\cos 0+e^{0} } = - \frac{1}{2}$
Hence, slope of tangent and normal at (0, 1) are 2 and $ - \frac{1}{2}$ respectively.
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives