for any $e$ prove that the difference between $f(x)$ and $f(x+e)$ is the same as the directional derivative of $f(x+te)$ in the direction of $e$

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Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.



So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.



I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.



And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$






real-analysis differential-equations derivatives proof-verification







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  • Yes, it is.$$
    – Lord Shark the Unknown
    Sep 11 at 4:21














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Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.



So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.



I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.



And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$






real-analysis differential-equations derivatives proof-verification







share|cite|improve this question























  • Yes, it is.$$
    – Lord Shark the Unknown
    Sep 11 at 4:21












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.



So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.



I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.



And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$






real-analysis differential-equations derivatives proof-verification







share|cite|improve this question















Let $f : Bbb R^n → Bbb R$ and $x ∈ Bbb R^n$. Suppose that for any unit vector $e ∈ Bbb R^n$, the directional derivative $nabla_e f(x + te)$ exists for all $t ∈ [0, 1]$. Prove that $f(x+e)−f(x)=nabla_e f(x+te)$ for some $t ∈ (0, 1)$.



So I think that this is basically the mean value theorem but with directional derivatives? But my knowledge of directional derivatives is limited and I am unsure of where to even begin on this one.



I know that $(x+te)∈(x,x+e)$ but that doesn't imply anything about $f(x+te)$.



And just writing stuff down I have: $nabla_e f(x+te)=fracdfdx_1(x_1+te_1), e_1+...+fracdfdx_n(x_n+te_n), e_n$






real-analysis differential-equations derivatives proof-verification




real-analysis differential-equations derivatives proof-verification






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edited Sep 11 at 16:15









Harry49

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5,5502929










asked Sep 11 at 4:19









Big_Tubbz

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234











  • Yes, it is.$$
    – Lord Shark the Unknown
    Sep 11 at 4:21
















  • Yes, it is.$$
    – Lord Shark the Unknown
    Sep 11 at 4:21















Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21




Yes, it is.$$
– Lord Shark the Unknown
Sep 11 at 4:21










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Note that $x$ and $e$ are fixed here. Consider the auxiliary function
$$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
Then
$$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
Now "put it all together".






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    Note that $x$ and $e$ are fixed here. Consider the auxiliary function
    $$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
    Then
    $$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
    Now "put it all together".






    share|cite|improve this answer
























      up vote
      1
      down vote













      Note that $x$ and $e$ are fixed here. Consider the auxiliary function
      $$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
      Then
      $$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
      Now "put it all together".






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        Note that $x$ and $e$ are fixed here. Consider the auxiliary function
        $$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
        Then
        $$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
        Now "put it all together".






        share|cite|improve this answer












        Note that $x$ and $e$ are fixed here. Consider the auxiliary function
        $$phi(t):=f(x+t e)qquad(0leq tleq 1) .$$
        Then
        $$phi'(t)=lim_hto0phi(t+h)-phi(t)over h=lim_hto0f(x+te +he)-f(x+te)over h=nabla_ef(x+te) .$$
        Now "put it all together".







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 11 at 18:49









        Christian Blatter

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