Solveeit Logo

Question

Question: Find \[\dfrac{{dy}}{{dx}}\] , when \[x = {e^t} - 3\] and \[y = {e^t} + 5\]...

Find dydx\dfrac{{dy}}{{dx}} , when x=et3x = {e^t} - 3 and y=et+5y = {e^t} + 5

Explanation

Solution

Hint : Derivation of a function is basically a measure of sensitivity to change of function value with change in the argument where argument refers to the input whose output is to be found. Derivatives are useful in finding the slope of an equation, maxima, and minima of a function when the slope is zero and is also used to check a function, whether it is increasing or decreasing.

Complete step-by-step answer :
In this question the given two functions have a variable tt , so we will differentiate both the equations as a function of tt and then we will find the desired equation.
Given the function
x=et3x = {e^t} - 3
Now since the function has the input as variable tt so we will differentiate the function with respect to tt , hence differentiating the function with respect to tt , we get

\dfrac{d}{{dt}}\left( x \right) = \dfrac{d}{{dt}}\left( {{e^t} - 3} \right) \\\ \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {{e^t}} \right) - \dfrac{d}{{dt}}\left( 3 \right) \\\ \dfrac{{dx}}{{dt}} = {e^t}\left\\{ {\because \dfrac{{d\left( {{e^t}} \right)}}{{dt}} = {e^t}} \right\\} \;

Hence we get dxdt=et(i)\dfrac{{dx}}{{dt}} = {e^t} - - (i)
Now for the function
y=et+5y = {e^t} + 5
Here again we can see the function has the input as variable tt so we will differentiate the function with respect to tt , hence differentiating the function with respect to tt , we get

\dfrac{d}{{dt}}\left( y \right) = \dfrac{d}{{dt}}\left( {{e^t} + 5} \right) \\\ \dfrac{{dy}}{{dt}} = \dfrac{{d\left( {{e^t}} \right)}}{{dt}} + \dfrac{{d\left( 5 \right)}}{{dt}} \\\ \dfrac{{dy}}{{dt}} = {e^t}\left\\{ {\because \dfrac{{d\left( {{e^t}} \right)}}{{dt}} = {e^t}} \right\\} \;

Hence we get dydt=et(ii)\dfrac{{dy}}{{dt}} = {e^t} - - (ii)
Now since we need to find the value of dydx\dfrac{{dy}}{{dx}} , hence we will divide equation (ii) by equation (i), we get
dydtdxdt=etet\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{{e^t}}}{{{e^t}}}
By solving

dydt×dtdx=etet dydx=1   \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} = \dfrac{{{e^t}}}{{{e^t}}} \\\ \dfrac{{dy}}{{dx}} = 1 \;

Hence the value of dydx=1\dfrac{{dy}}{{dx}} = 1
So, the correct answer is “1”.

Note : As a constant term does not contain any variables with them when they are differentiated, then their value is zero. Derivation of a function is represented in ab\dfrac{a}{b} , where aa is the function which is being differentiated and b its independent variable by which function is being differentiated written as dydx\dfrac{{dy}}{{dx}} where y is the function.