Question

If we have an expression \[\sin \left( x+y \right)=\log \left( x+y \right)\], then: \[\dfrac{dy}{dx}=\].

Answer 1

Hint: Write the given equation in terms of x, y. Now take the left-hand side and right-hand side to the power of the exponent. Now apply logarithmic properties to vanish the logarithm term. As we now have x, y linear equations in 2 variables.

Complete step-by-step solution -
Now differentiate with respect to x on both sides. Now differentiate with respect to x on both sides. Now you get the differential term \[\dfrac{dy}{dx}\] term is a variable and the remaining all are constants. Now you have a single variable linear equation. Now find the coefficient of the \[\dfrac{dy}{dx}\] variable on both sides of the equation. Subtract the term with a coefficient of the variable on the right-hand side. Now similarly find the constant values on both sides of the equation. Subtract the constant value of the left-hand side on both sides of the equation. Now you get an equation with variable terms on the left-hand side and constant terms on the right-hand side. Now find the coefficient term of the variable on the left-hand side. Divide with this coefficient on both sides of the equation. Now you have only the variable with coefficient 1 on the left-hand side and some constant on the right-hand side. So, this constant will be your result.
Linear Polynomials: If the degree of the polynomial is 1 then they are called linear polynomials. For example x + 1, x + 2, x + 3.
Degree of Polynomial:
The highest power of the variable in a polynomial is called its degree. For example \[{{x}^{2}}+4x+2\] has degree of 2, x + 1: degree of 1, \[{{x}^{3}}+1\]: degree of 3, 2 is a polynomial of degree 0.
Given equation in the question is written in the form of:
\[\Rightarrow \sin \left( x+y \right)=\log \left( x+y \right)\]
Now by taking e to the power (term) on both sides, we get:
\[\Rightarrow {{e}^{\sin \left( x+y \right)}}={{e}^{\log \left( x+y \right)}}\]
By diminishing log term by property: \[{{e}^{{{\log }_{e}}a}}=a\], we get
\[\Rightarrow \left( x+y \right)={{e}^{\sin \left( x+y \right)}}\] ------ (1)
By differentiating on both sides, we get the equation as:
Using the following formulas for differentiating the equation (1):
\[\Rightarrow \dfrac{d}{dx}x=1,\dfrac{d}{dx}{{e}^{x}}={{e}^{x}},\dfrac{d}{dx}\sin x=\cos x\]
By substituting these the equation turns into the form of:
\[\Rightarrow 1+\dfrac{dy}{dx}={{e}^{\sin \left( x+y \right)}}\cos \left( x+y \right).\left( 1+\dfrac{dy}{dx} \right)\]
By substituting the equation (1) back into this equation, we get:
\[\Rightarrow 1+\dfrac{dy}{dx}=\left( x+y \right)\cos \left( x+y \right).\left( 1+\dfrac{dy}{dx} \right)\]
By subtracting the term \[\left( x+y \right)\cos \left( x+y \right)\dfrac{dy}{dx}\] on both sides we get it as:
\[\Rightarrow 1+\dfrac{dy}{dx}-\left( x+y \right)\cos \left( x+y \right)\dfrac{dy}{dx}=\left( x+y \right)\cos \left( x+y \right)\]
By subtracting the term 1 on both sides of the above equation we get:
\[\Rightarrow \dfrac{dy}{dx}-\left( x+y \right)\cos \left( x+y \right)\dfrac{dy}{dx}=\left( x+y \right)\cos \left( x+y \right)-1\]
By taking \[\dfrac{dy}{dx}\] common on left-hand side of equation we get it as:
\[\Rightarrow \dfrac{dy}{dx}\left( 1-\left( x+y \right)\cos \left( x+y \right) \right)=-1\left( 1-\cos \left( x+y \right)\left( x+y \right) \right)\]
By canceling the common terms on both sides of the equation we get:
\[\Rightarrow \dfrac{dy}{dx}=-1\]
Therefore -1 is the \[\dfrac{dy}{dx}\] in the given equation.

Note: The idea of substituting equation (1) back after the differentiation step is very crucial as it is the main point which helps to reach the result. Generally, students forget to write 1 on the left-hand side while differentiating as their concentration will be on \[\dfrac{dy}{dx}\] they forget about 1. But this is the important term to cancel the term in the last step. So, don’t forget to consider “\[\dfrac{d}{dx}x=1\]”.