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Chapter 16: Problem 8

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given \(\ln (x+y)=k, k\) constant.

Short Answer

Expert verified
Question: Find the derivative of \(y\) with respect to \(x\) for the equation \(\ln(x+y)=k\). Answer: \(\frac{\mathrm{d}y}{\mathrm{d}x}=-1\)

Step by step solution

01

Differentiate implicitly with respect to x

To find \(\frac{\mathrm{d}y}{\mathrm{d}x}\), we will first differentiate both sides of the given equation with respect to \(x\). Using implicit differentiation, we have: $$ \frac{\mathrm{d}}{\mathrm{d}x}(\ln(x+y)) = \frac{\mathrm{d}}{\mathrm{d}x}(k) $$
02

Apply the chain rule

Since \(\ln(x+y)\) is a composite function, we will apply the chain rule while differentiating. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. For this exercise, the outer function is \(\ln(u)\) and the inner function is \(u = x + y\). Therefore, $$ \frac{\mathrm{d}\ln(u)}{\mathrm{d}x} \times \frac{\mathrm{d}(x+y)}{\mathrm{d}x} = 0 $$
03

Differentiate outer and inner functions

Proceed to differentiate \(\ln(u)\) with respect to \(u\) and \(x+y\) with respect to \(x\). We have: $$ \frac{1}{u}\frac{\mathrm{d}(x+y)}{\mathrm{d}x} = 0 $$ $$ \frac{1}{x+y}\left(\frac{\mathrm{d}x}{\mathrm{d}x} + \frac{\mathrm{d}y}{\mathrm{d}x}\right) = 0 $$
04

Simplify the equation

Simplify the equation by substituting for \(\frac{\mathrm{d}x}{\mathrm{d}x}\) and isolating \(\frac{\mathrm{d}y}{\mathrm{d}x}\): $$ \frac{1}{x+y}(1 + \frac{\mathrm{d}y}{\mathrm{d}x}) = 0 $$
05

Solve for dy/dx

Multiply both sides by \((x+y)\) and solve for \(\frac{\mathrm{d}y}{\mathrm{d}x}\): $$ 1+\frac{\mathrm{d}y}{\mathrm{d}x}=0 $$ $$ \frac{\mathrm{d}y}{\mathrm{d}x}=-1 $$ Thus, the derivative of \(y\) with respect to \(x\) is \(\frac{\mathrm{d}y}{\mathrm{d}x}=-1\).

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Most popular questions from this chapter

Find the derivative of the following functions using logarithmic differentiation: (a) \(a(t)=\left(1+t^{2}\right)^{3}\left(1-t^{2}\right)^{4} \sin ^{3} t\) (b) \(b(r)=\mathrm{e}^{-r} \sin ^{5} 2 r \cos ^{4} 3 r\) (c) \(K(p)=6 p \sqrt{\sin p}(\cos 2 p)^{1 / 3}\) (d) \(N(y)=\frac{(1+\sqrt{y})^{4}(6+7 y)^{3}}{\sqrt{1+y}}\) (e) \(x(t)=\frac{6}{(2+t)^{3}(1-t)^{4} \cos ^{3} t}\)

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given (a) \(x^{2}+x^{3}+y^{2}-y^{3}=1\) (b) \(2 x^{2}-y^{2}+3 x y-7 x-10 y=0\) (c) \(x y^{2}+\frac{y}{x}=\mathrm{e}^{x}\) (d) \(\ln (x y)-\sqrt{x}=\sqrt{y}\) (e) \(x \sin y+y^{2} \cos 2 x=y\)

Find the equation of the tangent to \(y=x \mathrm{e}^{x}\) where \(x=1\).

Find the equation of the normal to \(y=\ln x\) where \(x=2\).

Find \(\frac{\mathrm{dy}}{\mathrm{d} x}\) where \(y\) is given by (a) \(x \cos x\) (b) \(x e^{x^{k}}\) (c) \(\sin x \cos 2 x\) (d) \(x^{3} \mathrm{e}^{2 x}\) (e) \(x^{4} \sin 2 x\)
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