Differentiation on tangent on a point

Question

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

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

Answer 1

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

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 $$\ln w =x\ln x$$ Differentiate both sides w.r.t. $x$, $$\frac{1}{w}\cdot\frac{dw}{dx}=x\cdot\frac{1}{x}+\ln x\cdot 1$$ $$\frac{1}{x^x}\cdot \frac{dw}{dx}=1+\ln x$$ $$\frac{dw}{dx}=x^x(1+\ln x)\quad (3)$$

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 :)