Question

Find dy/dx of siny + x = logx

Answer 1

Answer:

siny + x = logx

diff. w r.t. x

d/dx siny + d/dx x= d/dx logx

cosy.dy/dx + 1 = 1/x

cosy.dy/dx = 1-x/x

dy/dx = 1-x/xcosy

Answer 2

Given:

siny + x = logx

To Find:

Find dy/dx

Solution:

The given equation is siny +x= logx before differentiating the given equation we should know the differentiation of sinx, x and logx which is cosx, 1 and 1/x respectively. Now differentiating the equation with respect to x,

$siny+x=logx\\cosy*\frac{dy}{dx} +1=\frac{1}{x}$

Now arranging the equation in terms of dy/dx ( we can change the value of cosy in terms of x using basic trigonometry but in the question, nothing is asked like that so we will leave it )

$cosy*\frac{dy}{dx} +1=\frac{1}{x} \\cosy*\frac{dy}{dx} =\frac{1-x}{x} \\\frac{dy}{dx} =\frac{1-x}{x*cosy}$

Hence, dy/dx of siny+x=logx is (1-x)/(x*cosy).