How do you find the derivative of $x = tan (x + y)$ $x = tan (x+y)$ ?

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How do you find the derivative of $x = tan (x + y)$ $x = tan (x+y)$ ?

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Answer 1 · 2016-07-12T07:10:07

$\frac{d y}{d x} = - \frac{x^{2}}{1 + x^{2}} .$ $dy/dx=-x^2/(1+x^2).$

Explanation:

$x = tan (x + y)$ $x=tan(x+y)$

Diff.ing, both sides w.r.t. $y$ $y$ , and keeping in mind the Chain Rule,

$\frac{d x}{d y} = \frac{d}{d y} tan (x + y) = ({sec}^{2} (x + y)) \frac{d}{d y} (x + y) = {sec}^{2} (x + y) \cdot (\frac{d x}{d y} + 1)$ $dx/dy=d/dytan(x+y)=(sec^2(x+y))d/dy(x+y)=sec^2(x+y)*(dx/dy+1)$

$\frac{d x}{d y} - ({sec}^{2} (x + y)) \frac{d x}{d y} = {sec}^{2} (x + y)$ $dx/dy-(sec^2(x+y))dx/dy=sec^2(x+y)$

${1 - {sec}^{2} (x + y)} \frac{d x}{d y} = {sec}^{2} (x + y)$ ${1-sec^2(x+y)}dx/dy=sec^2(x+y)$

Using, ${sec}^{2} θ = 1 + {tan}^{2} θ$ $sec^2theta=1+tan^2theta$ , we have,

$(- {tan}^{2} (x + y)) \frac{d x}{d y} = 1 + {tan}^{2} (x + y)$ $(-tan^2(x+y))dx/dy=1+tan^2(x+y)$

Knowing that, we have $tan (x + y) = x$ $tan(x+y)=x$ .

$- x^{2} \frac{d x}{d y} = 1 + x^{2}$ $-x^2dx/dy=1+x^2$ , giving, $\frac{d x}{d y} = - \frac{1 + x^{2}}{x^{2}}$ $dx/dy=-(1+x^2)/x^2$

Therefore, $\frac{d y}{d x} = - \frac{x^{2}}{1 + x^{2}} .$ $dy/dx=-x^2/(1+x^2).$

Method II

$x = tan (x + y)$ $x=tan(x+y)$

$\Rightarrow \frac{d}{d x} x = \frac{d}{d x} (tan (x + y))$ $rArr d/dxx=d/dx(tan(x+y))$

$\Rightarrow 1 = {sec}^{2} (x + y) {\frac{d}{d x} (x + y)} = {sec}^{2} (x + y) {1 + \frac{d y}{d x}} = (1 + x^{2}) {1 + \frac{d y}{d x}}$ $rArr 1=sec^2(x+y){d/dx(x+y)}=sec^2(x+y){1+dy/dx}=(1+x^2){1+dy/dx}$

$\Rightarrow 1 + \frac{d y}{d x} = \frac{1}{1 + x^{2}} \Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{2}} - 1 = \frac{1 - 1 - x^{2}}{1 + x^{2}}$ $rArr 1+dy/dx=1/(1+x^2) rArr dy/dx=1/(1+x^2)-1=(1-1-x^2)/(1+x^2)$

$\Rightarrow \frac{d y}{d x} = - \frac{x^{2}}{1 + x^{2}},$ $rArr dy/dx=-x^2/(1+x^2),$ as in Method I !

Method III

$x = tan (x + y)$ $x=tan(x+y)$

$arctan x = x + y \Rightarrow arctan x - x = y$ $arctanx=x+y rArr arctanx-x=y$

$\Rightarrow \frac{d y}{d x} = \frac{1}{1 + x^{2}} - 1 = - \frac{x^{2}}{1 + x^{2}}$ $rArr dy/dx=1/(1+x^2)-1 =-x^2/(1+x^2)$ , as derived before!

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How do you find the derivative of $x = tan (x + y)$ $x = tan (x+y)$ ?

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