# Differentiation on tangent on a point

Mathematics Asked by kk6 on December 18, 2020

$$text{tangent of } e^{x^2} + ln x + e^{xln x}$$

How to find the tangent of this curve at $$(1, e+1)$$ and differentiate it?

Let $$y=e^{x^2}+ln x+e^{xln x}$$.

Using the fact that $$e^{ln x}=x$$, we have $$y=e^{x^2}+ln x+x^x$$

Let us suppose for simplicity, $$u=e^{x^2}$$, $$v=ln x$$ and $$w=x^x$$. $$y=u+v+w$$ $$implies frac{dy}{dx}=frac{du}{dx}+frac{dv}{dx}+frac{dw}{dx}quad (*)$$

Differentiating $$u$$

To differentiate $$u=e^{x^2}$$, we let $$t=x^2$$. By Chain Rule, $$frac{du}{dx}=frac{du}{dt}cdot frac{dt}{dx}$$ $$frac{du}{dx}=frac{d}{dt}(e^t)cdot frac{d}{dx}(x^2)$$ $$frac{du}{dx}=2xe^{x^2}quad (1)$$

Differentiating $$v$$ is trivial as we know $$(ln x)'=1/x$$. So, $$frac{dv}{dx}=frac{1}{x}quad (2)$$

Differentiating $$w$$

For this, take $$ln$$ of both sides $$ln w =xln x$$ Differentiate both sides w.r.t. $$x$$, $$frac{1}{w}cdotfrac{dw}{dx}=xcdotfrac{1}{x}+ln xcdot 1$$ $$frac{1}{x^x}cdot frac{dw}{dx}=1+ln x$$ $$frac{dw}{dx}=x^x(1+ln x)quad (3)$$

Using $$(1)$$, $$(2)$$ and $$(3)$$ in $$(*)$$, we get $$frac{dy}{dx}=2xe^{x^2}+frac{1}{x}+x^x(1+ln x)$$

Thus the slope of $$y$$ at the point $$(1,e+1)$$ is obtained by putting $$x=1$$, $$left[frac{dy}{dx}right]_{x=1}=2e+1+1=2e+2$$

Thus, the tangent has slope $$m=2e+2$$ and lies on $$(x_0,y_0)=(1, e+1)$$. By point-slope form, $$y-y_0=m(x-x_0)$$ So, the equation of required tangent is $$y-e-1=(2e+2)(x-1)$$ $$(2e+2)x-y-(e-1)=0$$

Hope this helps :)

Answered by ultralegend5385 on December 18, 2020

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