What is the equation of the line tangent to $f (x) = x^{2} e^{x} - x e^{x^{2}}$ $f(x)=x^2e^x-xe^(x^2)$ at $x = 0$ $x=0$ ?

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Answer 1 · 2017-12-19T07:03:37

Let's see.

Given,

$f (x) = x^{2} e^{x} + x e^{x^{2}}$ $f(x)=x^2e^x+xe^(x^2)$

Differentiate it and find the slope $\frac{d y}{d x}$ $dy/dx$ .

Now find the $\frac{d y}{d x} (x = 0)$ $dy/dx(x=0)$

This slope is the slope of the tangent at that coordinate.

Now find the $f (0)$ $f(0)$ .

$:.x_1=0, y_1=f(0)$

$:.x_1=0, y_1=0$

Now use the equation of straight lines $rarr$

$y-y_1=dy/dx(x-x_1)$

$:.y-0=dy/dx(x=0)cdot(x-0)$

$:.y/x=dy/dx(x=0)$

Substitute the value of $dy/dx(x=0)$ and get the suitable equation of the tangent.

Hope it Helps:)

What is the equation of the line tangent to f(x)=x^2e^x-xe^(x^2) f(x)=x2ex−xex2 f(x)=x^2e^x-xe^(x^2) at x=0x=0 x=0?