# What is the y-intercept of the tangent to the curve y = e^x*sin x at the point where x = 0?

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### 2 Answers

Find the y-intercept of the tangent to the curve `y=e^xsinx ` at x=0:

The slope of the tangent line is found by computing the value of the function's first derivative at the given point, assuming the derivative exists. Then, with the point of the graph at the given x-value, you can write the equation of the tangent line and determine the y-intercept.

`f(x)=e^xsinx ==> f'(x)=e^xsinx+e^xcosx ` (using the product rule.)

Then `f'(0)=e^0sin0+e^0cos0=1 ` ; thus the slope of the tangent line is 1.

The function, and the tangent line, will both go through the point (0,f(0)) or (0,0) (since `f(0)=e^0sin0=0 ` .)

Thus the equation of the tangent line is y=x.

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The y-intercept of the tangent line is 0.

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The graph of the function and the tangent line:

Note that in the special case where you are asked to find the y-intercept of the tangent line to a graph of a function at x=0, it suffices to evaluate the function at x=0 **assuming the first derivative exists at x=0.** Here since the function is infinitely differentiable, all we needed to do was find f(0), since the tangent line by definition will include the same point.

However, given a function like y=|x|, if asked to find the y-intercept of the tangent line at x=0, the answer would be that there is no tangent line at x=0 as the first derivative fails to exist. So you must be careful when using "shortcuts".

The y-intercept of the tangent to the curve y = e^x*sin x at the point where x = 0 has to be determined.

For a curve y = f(x), the slope of the tangent to the curve at a point `(y_0, x_0)` is given by `f'(x_0)` . The equation of the tangent is `(y - y_0)/(x - x_0) = f'(x_0)` . The y-intercept can be determined by equating `x_0 = 0` .

Here, the y-intercept of the tangent to the curve at the point x = 0 is required. As the value of x is equal to 0, the y-coordinate of the point at which the tangent is drawn is the required value.

Solve y = e^0*sin 0,

y = 1*0 = 0

From the graph of the curve given below, it can be seen that a tangent drawn at x = 0 passes through the origin.

**The y-intercept of the tangent to the curve y = e^x*sin x at the point where x = 0 is 0.**